
Class TA3iO. 



£4: 



02? 



* ' - fl 

APPLIED MECHANICS 

A Treatise for the use of Students who have time to ujorh 

Experimented, Numerical, and Graphical exercises 

illustrating the subject 



BY 

JOHN PERRY, M.E, D.Sc, F.R.S. 

Whit. Sch., Assoc. M.Inst. C.E. 

professor of mechanics and mathematics in the royal college of science. 

south kensington ; vice-president of the physical society 

vice-president of the institution of electrical engineers 



IVITH S71 ILLUSTRATIONS 



NEW Y (3 K K 

D. VAN NO STRAND COMPANY 

23 MURRAY, AND 27 WAEEEN STS 
19U1 



^ 



^50 

P46 



'03 



• • • t • • < 



l^C« 



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PREFACE 



This book descvibes what lias for many years been the course 
of instruction in Applied Mechani<;s at the Finsbury Technical 
College. All mechanical and electrical engineering students 
in their first year had two lectures a week ; the substance of 
these lectures is here printed in the larger type. Mechanical 
engineers had three lectures a week in their second year ; the- 
substance of these lectures is here printed in small type. As 
I found our arrangement of hours per week to work fairly 
well, I £:ive it here : — . ; .■. 



Mathematics 

Graphics and Machine Drawing- 
Mechanics : Lectures ... ... 

Mechanical Laboratory 
IViechanics : Numerical Exercises 
Mechanism ... ... 

Wood and Iron Workshops ... 



Mechanical Engineers. Electrical Engineers 



1st j-ear. 



2nd year. 



1st year. 2ud year. 



Chemists and building trade students also attended in the, 
mechanical department, and the mechanical enginee^-iag 
students had courses of study in the physics and chemistry 
departments of the College. The Mechanics course incl^dodi 
\fork on the steam and gas engine not given in this book.-. 
When, after much experience in teaching at an English public 



IV PREFACE. 

school, in the Imperial College of Engineering, Japan, and 
other places, I ventured sixteen years ago to publish my 
method of teaching mechanics, it was met with some ridicule. 
Even without encouragement I was prepared to pursue the 
course which I had tested and found to be good, but I met 
with a great deal of encouragement from thoughtful men of 
about my own age. I had myself to scheme out and make 
drawings for every piece of apparatus for the laboratory, I 
knew of no collection of numerical and graphical exercises 
which were suitable for students, but gradually a collection 
was made from those given out in the lectures which seemed 
to me to be less objectionable, less academic, less misleading, 
than those hitherto available. I had difficulty in getting 
clever assistants trained in academic ways to sympathise 
with me. It was found in time that students took very eagerly 
to the quantitative experimental work, and that the whole 
system, faithfully followed, produced men whose knowledge was 
always ready for use in practical problems, and who knew the 
limits of usefulness of their knowledge. I am glad to say that 
more than twenty complete sets of the apparatus have been 
made and sent to various institutions by my workshop assistant, 
Mr. Shepherd. 

It would have made this book too large if I had included 
in it, as I should have liked to do, copies of the instructions 
which each student receives when he begins on a new piece of 
apparatus. 

Professor (now Sir Robert) Ball, at the Koyal College of 
Science, Dublin, started quantitative experimental mechanical 
work. He used the well-known frame of the late Professor 
Willis, which was taken to pieces and built up in new forms 
for fresh experiments. What I have done has been to carry 
out Professor Ball's idea, using a distinct piece of apparatus 
for each fresh kind of experiment. A student measures things 
for himself ; illustrates mechanical principles ; finds the limits 
to which the notions of the books as to friction and properties 
of materials are correct ; learns the use of squared paper, and 



PREFACE. T 

the accuracy of graphical methods of calculation ; and, above 
■all, really learns to think for himself. Professor Ewing, at 
Cambridge, has developed the ideas of Professor Ball to a far 
greater extent than what I have had opportunity for, and I 
know of no place in which a better engineering education can 
he obtained at the present time than at Cambriige. I am 
glad to think that a system begun under the Sciei ce and Art 
Department by Sir Robert Ball is now likely to be adopted 
generally in science classes. 

I am under great obligations to my assistant, Mr. G. A. 
Baxandall, who has been to great trouble in adding to the 
exercises, verifying answers, and correcting proofs. Professor 
Willis, D.Sc, has been kind enough to read through the 
proofs, and consequently I feel that there can be no im- 
portant mistake anywhere. 

I should like to think that, before a student begins the part 
in small type, he has worked through Thomson and Tait's 
«mall book on " Natural Philosophy," and that he has read the 
•early part of my book on " The Calculus for Engineers." 

JOHN PERRY. 

I6th July, 1897. 

Eoyal College of Science, 

London S.W. 



CONTENTS 



■CHAP. PAGE 

I. INTEODUCTOKT . . . . 1 

II. Tectoks : Relative Motion. . . . . . .29 

ni. Work and Energy. .38 

IV. Friction. . .' . . . . . . , . 56 

V. Efficiency . . .88 

VI. Machines: Special Cases .98 

VII. Elementary Analytical and Graphical Methods . . 120 
VIII. Examples in Graphical Statics ..... 151 

IX. Hydraulic Machines . 170 

X. Machinery in General 220 

XI. Ejnetic Energy . . 242 

XII. Materials Used in Construction 275 

XIII. Tension and Compression . . . . . . . 290 

XIV. Shear and Twist . . , . . . . . . 323 

XV. More Difficult Theory .... , , . 361 

XVI. Bending 381 

XVII. Strength at any Section of a Beam .... 393 
XVIII. Some Well-known Rules about Beams . . . . 410 

XIX. Diagrams of Bending Moment and Shearing Force . 427 

XX. More Difficult Cases of Bending of Beams . . . 441 

XXI. Bending and Crushing , . 457 

XXII. Metal Arches 478 

XXIII. Measurement of a Blow -.-.... 486 

XXIV. Fluids in Motion 505 

XXV. Periodic Motion . 54g 

XXVI. Mechanism . . . , , . . , ... 570 

XXVII. Centrifugal Force 597 

XXVIII. Springs gl3 

XXIX. Resilience of Springs g41 

XXX. Carriage Springs . g^g 



Appendix 



654 



Index . gg^- 



LIST OF TABLES. 



I. 

II. 
III. 

IV. 
V. 
VI. 

VII. 
VIII. 



IX. 

X. 

XI. 
XII. 

XIII. 

XIV. 

XV. 

XVI. 

xvn. 

XVIII. 

XIX. 

XX. 

XXI. 

XXII. 

XXIII. 

XXIV. 

XXV. 



Normal Pressure of Wind against Eoofs 157 

Moments of Inertia and the M and Tc of Kotating Bodies . . 251 

Young's Modulus (Wertheim) 302 

Factors of Safety for Different Materials and Loading . . 305 

Ultimate Stresses for Loads Eepeated a Great Number of Times 310 

Moment of Inertia and Strength Modulus of Various Sections . 

398-400 
Strength and Stiffness of Kectangular Beams Supported at the 
Ends and Loaded in the Middle 411 

Diagrams of Bending Moment, with Strength and Stiffness, of 
a Uniform Beam when Supported or Fixed and Loaded in 

Various Ways 414-16 

Breaking Stress of Iron and Timber Struts .... 468 
Values of the Constant B for Different Lengths of Struts and 
Different Materials 468 

Values of the Constant n for Struts of Different Sections . 469 
Relative Amplitudes for Different Frequencies in a Case of 

Forced Vibration 616 

Spring Materials Subjected to Bending 621 

Materials for Cylindric Spiral Springs 622 

Values of Torsional and Flexural Rigidity r.nd Axial Load for 

Various Sections of Wire used in Springs .... 
Relative Elongations of Various Springs for Same Axial Loads. 
Proof Load and Resilience of Various Springs .... 
Resilience of Spring Materials for Different Kinds of Loading . 
Useful Constants 
Moduli of Rigidity 



Moduli of Compressibility . 
Melting Points, Specific Gravity 
Logarithms .... 
Antilogarithms . 



Strength, 



635 
638 
640 
642 

654-5 
656 

. 657 
etc., of Materials 658-61 

662-3 

664-5 



Angles in Degrees and Radians, Sines, Tangents, etc. 



666 



Applied Mechanics. 



CHAPTER I. 

INTRODUCTO RT. 

1. The student of Applied Mechanics is supposed to have some 
acquaintance already with the principles of mechanics ; to be 
able to multiply and divide numbers and to use logarithms ; 
to have done a little practical geometry; to know a little 
algebra and the definitions of sine, cosine, and tangent of an 
angle ; and to have used squared paper. He is supposed to 
be working many numerical and graphical exercises; to be 
spending four hours a week at least in a mechanical laboratory ; 
to be learning about materials and tools in an iron and wood 
workshop; and to be getting acquainted with gearing and 
engineering appliances in a drawing-office and elsewhere. 

Unfortunately, many students are deceived as to their fitness 
to begin the study of Applied Mechanics, and we think it 
necessary in this introductory chapter to suggest some pre- 
paratory exercise work, and also to state certain definitions 
and facts which will be afterwards referred to, perhaps as if 
they were still unknown. 

When we think of what goes on under the name of teaching 
"we can almost forgive a man who uses a method of Ids own, 
however unscientific it may seem to be. Nevertheless, it is not 
easy to forgive men who, because they have found a study interest- 
ing themselves, make their students waste a term upon it, when 
only a few exercises are wanted — on what is sometimes called the 
scientific study of arithmetic, for example, or of mensuration. 

In our own suhject of Applied Mechanics there are teachers 
who spend most of the time on graphical statics, or the graph- 
ing of functions on squared paper, or the cursory examination of 
thousands of models of mechanical contrivances. One teacher seems 
to think that appHed mechanics is simply the study of kinematics 
And mechanisms ; another, that it is simply exercise work on pure 
mechanics ; another, that it is the breaking of specimens on a large 
testing machine ; another, that it is the trying to do in a school or 
college what can only he done in real engineering works ; another, 
that it is mere graphics ; another, that it is all calculus and no 



I APPLIED MECUAMCS. 

graphics; another, that it is all shading- and coloming and the 
production of j^retty pictures without centre lines or dimensions. 
Probably the gxeatest mistake is that of wasting time in a school 
in giving the information that one cannot help picking up in one's 
ordinary practical work after lea^dng scbool. 

We believe that the principles which an engineer really 
recollects and keeps ready for mental use are very few. Bv 
means of lectures, models, drawing-office and laboratory, and 
numerical exercise work, we show a man how these 'simple 
principles enter, in ciuiously different-looking shapes, into his 
engineering practice. We give him the use of all the necessary 
methods of study, and we send him out into practical life pre"- 
pared to study things for himself. We ought to recogiiiso the 
fact that his real study of bis profession is not at school or college. 
We ought to teach him. how to learn for himself. Any child can 
state Newton's second law of motion, and the other half-dozen 
all-important principles of mechanics, so as to get full marks in an 
examination paper; the engineer knows that the phenomena he 
deals with are exceedingly complex, and that only a long ex- 
j)erience will enable him to utilise the so easily stated principles. 
Schools and colleges are the places in which men ought to learn 
the uses of all mental tools ; they are sure to specialise afterwards, 
but in the meantime we ought to give them plenty of tools to 
choose from. The average student cannot take in more than the 
elementary principles , the best students need not take in more. 

2. The most impoitant lesson for a beginner, however lie 
may have studied mathematics and mechanics, and however able 
he may be as a mathematician, is this — that he must not go on 
merely assuming that he knows how to do things ; he must know 
things by actual trial and not mere hearsay. He must actually 
calculate certain numerical results ; he must actually illustrate 
principles with laboratory apparatus ; and, if there is a school 
workshop, he must get to know the properties of materials by 
chipping and filing and paring and planing and turning. It 
is just the same as in one's after-school work. There is no 
great mechanical engineer who has not himself worked like 
a workman with other workmen, and got to understand men 
and things by actual contact with them. The man who shirks 
the following exercises and laboratory work will lose a great 
deal more than he is aware of. 

Teachers will notice that things requiring even a little pre- 
paration more than other things will gradually become neglected. 
Therefore, let it be part of the daily work for every student to use 
logarithms, drawing instruments in graphical exercise work, and 
squared paper. In the drawing-office, blue prints ought to be 
made by some student or other every day ; the planimeter ought 



APPLIED MECHANICS 3 

to be used every day, and some student ought to "be resetting his 
drawing-pens and other instruments every day. It ought to he a 
rule that all ajiparatus must always he ready for use, and that it is 
always in use. Teachers can arrange their own work in such a 
way that they cannot help seeing every day how the practical work 
of students is being done. When we find our system to be going 
with clockwork regularity and we feel no worry, we ought to 
believe that some change is necessary. If we find that the students 
are not absorbed in their work, we must understand that we 
teachers are in fault. 

3. Students cannot spend too much time in multiplj'ing, 
factoring, and simplifying algebraical and trigonometrical 
expressions. These are our tools, and we must get familiar 
with them. We may easily spend too much time in studying 
roots of equations, permutations and combinations, etc., and 
in the solution of triangles ; and therefore, if it is possible, we 
try to learn all our mathematics, mechanics, physics, and 
chemistry from teachers who are engineers. What acquaint- 
ance with these subjects we have, ought to be a real knowledge, 
not the glib pretence which suffices for examinations ; it must 
not be something apart from our life and work. To effect this 
object we must work many numerical and graphical exercises, 
and try to conquer our contempt for simple laboratory experi- 
ments, and illustrate the forty-seventh proposition of the Firot 
Book of Euclid by actually drawing some right-angled triangles 
and measuring their sides — to illustrate rules about triangles 
by actual measurement. In this way we learn much more 
than the ordinary geometrician knows ; among other things, 
we obtain a valuable knowledge of the errors we are likely to 
make in graphical calculation. 

4. Every student is supposed to be able to calculate the 
values of any algebraical or trigonometrical expression when 
numerical values are given. He ought to be able, when given 
at random such an expression as 

w = a 3 J-V2 ^ ^m^ -j- n'^ a loge b . cos. B, 
to be able to calculate iv when be knows the values of a, b, m^ 
n, and 0. He ought to be able to use logarithms in multiplica- 
tion and division and extraction of roots ; to know all the usual 
mathematical. symbols, and how it is sometimes convenient to 
use ^ya and sometimes a^. It is pedantic to say that a 
man must not ^se a formula unless he is able to prove its 
truth. It is usilally of great help in learning to prove a 



4 APPLIED MECHANICS. 

formula to have previously used the formula and know the 
value and meaning of what we are to prove. A living 
Northern professor of great eminence has declared that a boy 
ought not to be allowed to use logarithms until he is able 
to calculate them ; he has not said that a boy ought not to 
use a watch or wear a coat until he is able to make them. 

EXERCISES. 

1. Find 4-326 x 0-003457 to four significant figures, leaving out all 
unnecessary figures in the work. Find 0-01584 4- 2104 to four signifi- 
cant figm-es. Also do these using logarithms. Find log-g 7. Calculate 

5 2-43^ 3-0-246^ 042 0-4^6^ /y/246-3, 31-01* x 002641T . 

Ans., 001495, 0-007529, 1-94591, 49'9o, 0-7632, 0-2211, 3-008, o-872. 

2. If 7n =z {a^ -\- 2 cfib + s - '345)1 ^ (^2 b'^f 

find m\ia = 0*504, b = 0-309, s = 1'667. 

Ans., 1-4-53 x 10^. 

3. What errors are there in assuming 

(1 + a)« = 1 -\- im 
to be true in (1-001)3 = 1-003, (l-Ol)* = 10033. 

(0-99)2 = (1 - -01)2 = 1 - 02 = -98. 

i- = ^ = (1 - -on -1 = 1 + -01 = 101. "• 

•99 1 _ -01 ^ ^ 

— = (I 4- -on ~^ 

yi-oi 

^99 = yiOO (1 - -01) = 10 (1 - -01)^ = 10 (1 - -005). 

= 9-95 ? 
The above answers are very nearly correct ; the student is expected to 
find the correct answers. 

4. How much error is there in the assumptions 

L+^ = 1 + „ _ ^, (1 + „) (1 + ^) = 1 + a + ;3, 

when a = -01 /8 = -01, a = - -003 $= - '005 ? 

A)is., Xo error : -01 per cent., -004 per cent., -0015 per cent. 

5. If d is the diameter of the bore or the " calibre " of a gun, it is 
usually assumed that the weight of the gun is proportional to d^, and 
that the thickness of armour which its projectile will pierce is propor- 
tional to d. If an 8-inch gun weighs 14 tons and can pierce 11 inches of 
armoirr, what thickness will be pierced by a 10-inch gun, and what is the 
weight of the gun? Ans., 13-75 inches, 27*34 tons. 

5. The linear expansion of bodies by heat is practically 
proportional to the rise of temperature. The values of a, the 
co-efficient for linear expansion (the fractional increase in 
length for a rise in temperature of 1° Centigrade), are the 



3 ^ _ =(1 + -01) ^r= (1 - -0033) = -9967. 



APPLIED MECHANICS. 

following numbers divided by 10^ : — Aluminium, 2-34 ; copper, 
1-79; gold, 145; iron, 1-2; lead, 2*95 ; platinum, 0-9 ; silver, 
1-94; tin, 2-27 ; zinc, 2-9; brass (71 copper to 29 zinc), 1-87; 
bronze (86 copper to 10 tin to 4 zinc), 1*8; German silver, 
1'8; steel, I'll; brick, 0-5; glass, 0-9; granite, 0*9; sand- 
stone, 1-2; slate, 1*04; boxwood (across the fibre), 6-1; box- 
wood (along the fibre), 0'3; oak (across), 5*4; oak (along), 
0-5 ; pine (across), 3-4; pine (along), 0*5. 

The co-eiEcient, k, of cubical expansion is three times the 
co-efficient of linear expansion, because (1 + a)^ = 1 + 3 a 
is practically correct for these small values of a. The average 
values of k between 0° and 100° C. are the following numbers 
divided by 10^ : — Alcohol, 1*26 ; mercury, 0*18 ; olive oil, 0*8; 
petroleum, 1*04 ; pure water, 0*43 ; sea- water, 0-5. 

The student is supposed to have worked many exercises 
like the following ones : — 

1. Steel rails of 0° C. have an aggregate length of 1 mile. What is 
the length at 33.° C. ? Ans., 1 mile 24-2 inches. 

2. A ring of wrought iron has an inside diameter of 5 feet when at a 
temperature of 970° C. "What is the diameter at 0° C. ? Ans., 4'9 feet. 

3. A cylindric plug of copper just fits into a hole 4 inches diameter in 
a piece of cast iron. After heating the mass to 1,240° C, by how much 
is the diameter of the hole too small for the plug? Ans., -0293 inch. 

4. A bar of iron is 70 centimetres long at 0" 0. What is its length in 
boiling water (100° C.) ? What is its length at 50° C. ? 

Ans., 70-079 centimetres, 70-039 centimetres. 

5. Two rods — one of copper, the other of iron — measure 98 centimetres 
each at 0° C. What is the difference in their lengths at 57° C. ? 

Ans., -027 centimetre. 

6. Bars of wrought iron, each 3*4 metres long, are laid down at a 
temperature of 10° C. What space is left between every two if they are 
intended to close up completely at 40° 0. ? Ans., 1*26 millimetres. 

7. A wrought-iron connecting-rod is 12 feet long at 10° C. What is 
the increase of length at 80° 0. ? Ans., 0-121 inch. 

8. A wrought-iron Cornish boiler is 33 feet long; the shell is at 0° C, 
the flue at 100° C. What would the difference of the lengths be if the flue 
were not prevented from expansion ? Ans., 0-475 inch. 

9. A steel pimip rod is 1,000 feet long. What is its change of length 
for a change of 10° C. Ans., 1-44 inch, 

10. In a thermometer -01 cubic inch of mercury at 10° C. is raised to 
15° C, and rises 1 inch in the tube. What is the cross-section of the tube ? 

Ans., 9 X 10~^ square inch. 

11. The volume of a lump of iron being 5 cubic feet at 10° C, find its 
volume at 80° 0. Ans., 5-0126 cubic feet. 

6. A student's knowledge of mathematics ought to be such 
that he can work out for himself all the rules given in such an 



6 APPLIED MECHANICS. 

excellent book on mensuration as that of Professor A. Lodge. 
The thorough study of such a book is one of several ways 
which may be recommended of getthig familiar with mathe- 
matical principles. But nobody's Hfe is long enough to use 
all these ways, and, besides, unnecessary study leads to dul- 
ness. Hence, if a student has taken some other way, he need 
not be alarmed at his ignorance of the more complex rules in 
mensuration ; he may feel absolutely certain that he can work 
out such rules for himself, given time and necessity. He will 
study the more complex rules, such as prismoidal formulae, if 
he needs to use them practically, not otherwise. The following 
rules are in constant use and must be familiar to the student, 
whether or not he knows the reasons for them. If he is 
familiar with the rules and does not anxiously search for the 
reasons for them, he lacks the necessary spirit of the practical 



RULES IN MENSURATION. 

An area is found in square inches if all the dimensions are 
given in inches. It is found in square feet if all the dimen- 
sions are given in feet. 

Area of a ^;<:6ra?^e?o^ra??i. — Multiply the length of one side 
Dy the perpendicular distance from the opposite side. 

The centre of gravity of a parallelogram is at the point of 
intersection of its diagonals. 

Draw a right-angled triangle ; measure very accurately the 

lengths of the sides. You will find that, no matter what scale 

of measurement you use, the square of the length of the 

bypothenuse is equal to the sum of the squares of the lengths 

of the other two sides. 

Area of a triangle. — Any side multiplied 
by its perpendicular distance from the oppo- 
site corner and divided by two. 

The eeyitre of gravity, or, rather, the centre 
of area, of a triangle is found by joining 
(Fig. 1) any corner, a, with the middle 
point, D, of the opposite side, b c, and making d g one-third 
of D A. G is the centre of gravity. 

Area of an irregular figure. — Divide into triangles, and 

add the areas of the triangles together. 

■ - 
Circumference of a circle. — Multiply the diameter by 

3-1416. 




APPLIED MECHANICS. 7 

Arc of a circle. — From eight times the chord of half the 
arc subtract the chord of the whole arc; one-third of the 
remainder will give the length of the arc, nearly. 

Area of a trapezium. — Half the sum of the parallel sides 
multiplied by the perpendicular distance between ^hem. 

Area of a circle. — Square the radius, and multiply by 
3'1416 ; or square the diameter, and multiply by 0-7854. 

Area of a sector of a circle. — Multiply half the length of 
the arc by the radius of the circle. 

Area of a segment of a circle. — Find the area of the sector 
having the same arc, and the area of the triangle formed by 
the chord of the segment and the two radii of the sector. 
Take the sum or difference of these areas as the segment is 
greater or less than a semicircle. 

Otherwise, for an approximate answer : — Divide the cube 
of the height of the segment by twice the chord, and add the 
quotient to two-thirds of the product of the chord and height 
of the segment. When the segment is greater than a semi- 
circle, subtract the area of the remaining segment from the 
area of the circle. 

The areas of curves may be found by Simpson's rule. — 
Divide the area into any even number of parts by an odd 
number of equidistant parallel lines or ordinates, the first and 
last touching the bounding curve. Take the sum of the 
extreme ordinates (in many cases each of the extreme 
ordinates is of no length), four times the sum of the even 
ordinates, and twice the sum of the odd ordinates (omitting 
the first and last) ; multiply the total sum by one-third of the 
distance between any two successive ordinates. 

The ordinary rule for an indicator diagram is : — Draw 
lines at right angles to the atmospheric line, touching the 
extreme ends of the diagram. Divide the distance between 
them into ten equal parts (a parallel ruler with ten pieces is 
sometimes supplied), and at the middle of each part draw a 
line at right angles to the atmospheric line. Measure the 
breadth of the diagram on each of these ten lines, and take 
one-tenth of their sum. This gives the average breadth, and 
represents the average pressure to scale. The better plan is 
to find the area by a planimeter. The earnest student will 
practise the use of the planimeter, finding its error by tests on 
rectangles and circles. The average breadth of a diagram is 
used in many ways. 



8 



APPLIED MECHANICS. 



Exercise for Advanced Students. — Prove that Simpson's 
rule gives the area correctly if the arcs of curve between the 
odd ordinates follow, each of them, any such law as 

y=: a -\- bx -\- cx^ . 

Surface of a sphere. — Multiply the diameter by the 
circumference. 

Surface of a cylinder. — Multiply the circumference by the 
length, and add the areas of the two ends. 

Surface of a right circular cone. — Multiply half the 
circumference of the base by the slant side, and add the area 
of the base. 

Lateral surface of the frustum of a right cone. — Multiply 
the slant side by the circumference of the section equidistant 
from its parallel faces. 

Area of an elli-pse. — Multiply the product of the major and 
minor axes by "TSo^. 

The areas of two similar figures are as the squares of their 
like dimensions. The volumes are as the cubes of their like 
dimensions. 

The cuhic content of a body is calculated in cubic inches if 
all the dimensions are given in inches ; in cubic feet if all the 
dimensions are given in feet. 

Cubic content of a plate. — Multiply area of plate by its 
thickness. 

Cubic content of a sphere. — Cube the diameter, and multiply 
by -5280. 

Cubic content of the segment of a spAere. ^Subtract twic<" 






Fig. 2. 



APPLIED MECHANICS. 



9 



the height of the segment from three times the diameter of the 
sphere ; multiply the remainder by the square of the height, 
and this product by •5236. 

The cubic content and surface of a sphere are each two- 
thirds of that of the cylindric vessel which just encloses it. 

Cubic content of any prismatic body (Fig. 2). — Multiply 
the area of the bd,se by the perpendicular height. This will 
give the same product as, Multiply the area of cross section by 
length along the axis of the prism. (The axis of a prismatic 
body goes from centre of gravity of base to centre of gravity 
of top.) The centre of gravity of a prismatic body is half-way 
along the axis. 

Cubic content of any pyramidal or conical body (Fig. 3). 






Fig. 3. 

— Multiply the area of the base by one-third of the perpen- 
dicular height. 

Centre of gravity is one-quarter of the way along the axis 
from the base. (The axis of any such body joins the centre of 
area of base with the vertex.) 

The cubic content of the rim of a wheel is found by multi- 
plying the area of a cross section by the circumference of the 
circle which passes through the centres of gravity of the cross 
sections. 

The weight of a cubic inch of each of the following mate- 
rials is given in lbs. : — Cast iron, -26; wrought iron, '28; steel, 
28 j brass, -3 ; copper, -32 ; bronze, '3 ; lead, "4; tin, '27 ; zinc, 
•26. Hence to find the weight of a l3ody of cast iron or any 
other of these substances, find the volume in cubic inches and 
multiply by one of the above numbers. 

Yery often it is only the approximate weight that is 
wanted, so that a moulder may know how much metal to melt, 
or for other purposes. Now, suppose we want the approxi- 
mate weight of a cast-iron beam. Find roughly the average 
section and get its area in square inches, multiply by the length 



10 APPLIED MECHANICS, 

in inches, add to this the cubic content of any little gusset 
plates or other excrescences, multiply by '26 and we have the 
weight in pounds. 

The specific gravity of a substance means its weight as 
compared with the weight of the same bulk of water. Now, 
it is known that a cubic foot of water weighs very nearly 1,000 
ounces, or rather 62-3 lbs. The specific gravity of a brick 
varies from 2 to 2-167, and therefore the weight of a cubic foot 
of brick varies from 2x1,000 or 2,000 to 2-167x1,000, or 
2,167 ounces. 

We see, then, that from a table of specific gravities we can 
get the weight of a cubic foot of a substance, and therefore if 
we know the cubic content of a body formed of this substance 
we can calculate its weight. 

\^arious plans for saving labour in calculation suggest 
themselves to people working at any particular trade. For 
instance, if a pattern has no prints for cores, the weight of the 
pattern bears nearly the same proportion to the weight of the 
casting as the weight of a cubic inch of the wood bears to the 
weight of a cubic inch of cast iron. This is not always a con- 
venient rule, because the pattern is a little larger than the 
casting, and the density of wood alters as it dries. 

The area of an irregular figure may be obtained approxi- 
mately by cutting it out of a uniform sheet of cardboard and 
weighing it. Now cut out a rectangle or square whose area it 
is easy to calculate. Weigh this also. The areas are in the 
same proportion as the weights. 

The area of cross section of a fine luire in square inches can 
be determined with some accuracy by weighing a considerable 
length of the wire, dividing by the weight of the material per 
cubic inch, and dividing by the length of the wire in inches. 
EXEE0ISE3. 

1 . Find the number of revolutions per mile made by a rolling wheel 
4| feet diameter. Ans., 373. 

2. Find the area and circumference of a circle of 4 inches radius. 
Find the circumference and diameter of a circle whose area is 20 square 
inches. Ans., 50-27 square inches, 25-13 inches; 15-85 inches, 5-046 inches. 

3. Find the area of a j)arailelogram whose adjacent sides are 50 and 
30 feet and the angle between them 65°. Ans., 1359-45 square feet. 

4. Find the area of a sector of n circle, radius 4 in.ches, angle 50°. 

A71S., 6-982 square inches. 

5. Find the area of the segment of a circle, chord 20 inches, height 
3 inches. If this were a parabolic segment, its area would t;e |rds of 
chord multiplied by height. Ans., 40-6 square inches. 



APPLIED MECHANICS. 11 

6. Ordinates of a curve, 1'5 inclies apart, are 2 "30, 2'35, 2-46, 2*57, 
i-42, 2-21, 2-10, Find the area "between tlie first and last by Simpson's 
rule. Test your answer by drawing the curve and using a planimeter. 

Ans., 21-34. 

7. Find the area of the surface of a sphere, its radius being 8 inches. 

Ans., 804*2 square inches. 

8. A boiler has 300 tubes 8 feet long, 3 inches diameter. What is the 
total cross-sectional area ? What is the area of tube-heating surface ? 

Ans., 2,121 square inches, 1,885 square feet. 

9. Right circular cone, base 3 inches radius, height 8 inches ; find its 
curved surface and volume. Ans., 80*52 square inches, 75'41 cubic inches. 

10. Segment of sphere, height 4 inches, diameter of base 15 inches : 
find volume. Ans., 309*9 cubic inches. 

11. Find by Simpson's rule the number of cubic feet in a log of 
timber 36 feet long, the cross- sections at intervals of 6 feet being 8*20, 
5*68, 4*04, 2*92, 2*16, 1*54, 1*02 square feet. Ans., 124*36. 

12. A railway cutting is ^ mile long; the thirteen cross-sectional 
areas, measured at equal intervals of 220 feet, are respectively 280, 462, 
594, 685, 757, 742, 500, 346, 320, 418, 512, 626, 560 square feet. Find 
the volume by Simpson's rule. Ans., 52,480 cubic yards. 

13. Two models of terrestrial globes ; the areas of Africa are in the 
ratio 3 to 2. What is the ratio of the diameters of the globes ? What is 
the ratio of their volumes ? Ans., 1*225, 1*837. 

14. Find the siirface and volume of a sphere of 2*361 inches radius. 
If it is of cast iron, what is its weight ? Find the weight of a segment 
of this sphere 1 inch in height. 

Ans., 70*1 square inches, 5512 cubic inches ; 14*44 lbs., 1-67 lb. 

15. Find the volume and weight of the rim of a cast-iron wheel ; 
section circular, outside and inside radii 20 feet and 18 feet 6 inches. 

Am., 213*7 cubic feet, 42*87 tons. 

16. Find the circumference and area of a circle whose radius is 
5 inches. This is the base of a cylinder of 11 inches height. ^Vhat is 
its volume ? Find its curved surface. If made of cast iron, what is its 
weight ? What is the area of a section making 70° with the axis ? 
(A X cos. 20° is the area of the circular base which is its projection.) 

Ans., 31*42 inches, 78*54 square inches, 863*9 cubic inches, 2*4 square 
feet, 226*4 lbs., 83*58 square inches. 

17. A cone on an elliptic base, whose principal diameters are 12 inches 
and 8 inches, is 20 inches in vertical height. What is its volume ? What 
is its weight if made of cast iron ? Ans., 502*7 cubic inches, 131*7 lbs. 

18. A plate of wrought iron ^^ inch thick weighs 0*6 lb. What is its 
area? Ans., 34*3 square inches. 

19. A disc of zinc 10 inches diameter, with a hole of 2 inches diameter 
in it, weighs 0*2 lb. Find its thickness. Ans., 0*0106 inch. 

20. 4 lbs. of copper is drawn into wire 014 inch diameter. Find its 
length. Ans., 67-66 feet. 

21. A piece of round copper wire 100 feet long weighs 5 lbs. What is 
its diameter ? Ans., '128 inch. 

22. A spherical shell 10 inches outside and 8 inches inside diameter, 
of cast iron : what is its weight ? Ans., 66*45 lbs. 



12 APPLIED MECHANICS. 

23. A rolled girder of wrougiit iron is 12 inches outside depth, flanges 
are 6 inches by J inch, web is ^ inch thick : what is its weight if 18 feet 

lo^S'^ ^/.s, b95-6 lbs. 

24. Cylindric boiler 30 feet long, 7 feet diameter. Two cylindric flues 
each 2 feet 6 inches diameter: what is the area of the plates? Plate 
everywhere ^ inch thick : what is its weight ? 

When in this boiler water covers the flues and its level is a quarter 
of the diameter of the shell from the top, what are the volumes of water 
and steam P ^^^^ l^;^g9 ^^^^.^^^ ^^^^^ jq.^ ^^^^^ ^-8 to 1. 

25. A circular plate of lead 2 inches thick, 8 inches diameter, is con- 
verted into spherical shot of the same density, each of -075 inch radius. 
How many shot does it make ? Ans., 56,889. 

26. Walking from the centre towards the end of one span of a lattice 
girder railway bridge, I count 10 wrought iron bars of rectangular section, 
each 14 feet long ; the cross section of the first is 5 inches x f inch. If 
the widths increase by i inch and the thicknesses by -^-^ inch, find the 
total weight of bars. Ans., 1-08 ton. 

7. The idea of velocity involves two things — the direction 
and the speed. When the direction does not alter, we speak of 
the speed as if it were the whole idea. Find the time in seconds 
taken by a body to traverse a certain distance measured in 
feet. This distance divided by the time is called the average 
velocity. Thus, if a railway train moves through 200 feet in 
4 seconds, its average velocity during this time is 200 -i- 4, or 
50 feet per second. If we find, with careful measuring 
instruments, that it moves through 20 feet in -4 second, or 
through 2 feet in -04 second, the velocity is 20 -f -4, or 
2 -f- '04, or 50 feet per second. It is important to remember 
that tlie velocity may be always changing during an interval 
of time, however short. To get the velocity at any instant, 
we must make very exact measurements of the time taken to 
pass over a very short distance, and even this will only give us 
the average velocity during this short time. But if we make 
a number of measurements, using shorter and shorter periods of 
time, the average velocity becomes more and more nearly the 
velocity which we want. Thus, at 10 o'clock, a man in a 
railway train making a careful measurement finds that tlie 
train passes over 200 feet in the next 4 seconds. He finds 
the average speed for 4 seconds after 10 o'clock to be 200 -f- 4, 
or 50 feet per second. Another man finds that it passes over 
100-4 feet in the two seconds after 10 o'clock, and finds during 
these two seconds an average velocity of 100-4 -4- 2, or 50*2 
feet per second. Another man finds 50-25 feet passed over in 



APPLIED MECHANICS. 13 

one second after 10 o'clock, which shows a velocity of 50*25 
feet per second. Another man finds 25-132 feet passed over 
in half a second after 10 o'clock, and finds 2o*132 -^ 0-5, or 
50-264 feet per second. Another man finds 12*567 feet in a 
quarter second after 10 o'clock, and his observation gives 
50-268 feet per second, and so on. It is evident that the 
values given by these various observations are approaching the 
real value of the velocity at 10 o'clock. 
Tabulating these results, we have : — 



Intervals of Time in Seconds 
after 10 o'clock. 


Average Velocity in Feet per 
Second. 


4 
2 

1 

i 


5000 

50-20 

50-25 

50-264 

50-268 



Plot the two sets of numbers on squared paper, and draw 
a curve through the points so found. Produce the curve, and 
we have the means of finding the average velocity for an 
infinitely small interval of time after 10 o'clock. This is the 
required velocity. 

8. Acceleration. — This is the time rate of change of th& 
velocity of a body. Thus it is known that the velocity of d 
body falling freely in London — 

At the end of one second is 32*2 feet per second. 
„ „ two seconds is 64*4: „ ,, 
„ „ three „ 96-6 „ „ 

„ four „ 128-8 
and we see that there is an increase to the velocity of 32-2 
every second. The acceleration in this case is always of the 
same amount — hence we call it uniform acceleration, and sav 
it is 32-2 feet per second per second. 

EXEECISES. 

1. One mile per hour; also one knot; convert each of these into 
feet per minute and feet per second. Ans., 88, 1-467 ; 101-3, 1-689. 

2. A torpedo-catcher travels at 32 knots ; convert this into Enp-lish 
miles per hour. j^^s., 36-85. 

3. Prove that 60 miles per hour means -0268 kilometres* per second. 

* See tables on page 654. 



14 APPLIED MECHANICS. 

4. An eccentric disc is 10 inches diameter ; the shaft makes 300 
revolutions per minute. What is the rubbing velocity on the straps in 
feet per second ? Ans., 1S09. 

5. In running a race of 1 mile long, A. beats B by 100 yards, and B 
beats C by 90 yards. By how many does A beat C ? 

Ans., 185 yards, nearly, 

6. Ten miles per hour; state this in feet per second and in centimetres 
per second. Ans., 14|; 447. 

7. An acceleration of 32'2 feet per second per second ; state this in 
miles per hour per second ; state it in centimetres per second per second. 

Ans., 21-96 ; 981-4. 

8. 200 gallons of water per minute ; how many pounds per second ? 
How many cubic feet per second ? Afis., 33-3 ; -535. 

9. A round pipe 6 inches diameter has 30 gallons per second flowing 
through it. What is the velocity ? If the diameter becomes 10 inches, 
what is the velocity ? 

Calculate in the two cases the kinetic energy of one pound of water, 
this being the square of the velocity divided by 64-4. 

Ans., 24 '5 feet per second ; 8-8 feet per second ; 9*3 foot-pounds ; r2 
foot-pound. 

9. Example. — Two fine wires are 10 feet apart; a bullet 
breaks them both. The breaking of each wire causes an 
electric spark to make a mark underneath a fixed platinum 
pointer on a revolving drum. If the drum is 4 feet in 
diameter, and revolves at 300 revolutions per minute, and 
when the drum is at rest the spark-marks are found to be 1'32 
feet asunder on the curved surface, assuming that the intervals 
of time between the breaking of the wires and making the 
marks were the same, find the time between the breaking of 
the wires, and find the velocity of the bullet. The surface 

velocity is ^n. , or 62-83 feet per second; 1*32 divided 

by this gives '02101 seconds ; dividing this into 10 feet gives 
476 feet per second as the velocity of the bullet. 

Exercise. — In some gun experiments screens 150 feet apart 
were cut by a bullet at the following times (in seconds), 
counting from the time of cutting the first screen : — 0, 0-0666, 
0-1343, 0-2031, 0-2729, 0-3439, 0-4159. Find the average 
velocity between every two successive screens. 

Ans., 2,252, 2,216, 2,180, 2,149, 2,113, and 2,083 feet per 
second. 

10. Speed or velocity is a rate— the rate of increase of space with 
regard to time. If a body has passed through the space s at the time t, 
and if it goes over the additional space ds in the additional time of 

St, then -^ is the average velocity. Observe that 5s is one symbol ; 
St 



APPLIED MECHANICS. 



15 



symbol ~ means 



it does not mean a quantity called S multiplied by a quantity called s. 

"When we imagine S;^ to get smaller and smaller without limit, the 

average velocity becomes what we call the real velocity, and we 

clft 
indicate it by the symbol -j- . AVhen, instead of space and tinie, 

we have other quantities, we generally use y for the dependent 

And X for the independent variable, and the 

*' the rate of increase of y with regard to .^•." Thus, if y is the 
ordinate of a curve and x is the abscissa at the point p (Fig. 
4), and q r =: 5y, p r = 5^-, then 5y/5a; is the tangent of the angle 
Qvx. As 8^ and Sy become smaller and smaller, so that ^yjdx is 
to be written dyjdx, then dylclx is evidently tangent of the angle 
which the tangent at p makes with the axis of x. We usually 
call this the slope of the curve at the point p. 

If y is any quantity that depends upon another, x, 
plot the values of x and y to 
scale on a sheet of squared 
paper, the curve shows by 
its slope everywhere the rate 
of increase of y with regard 
to X. If we know the alge- 
braic law connecting y and 
X, we can find this rate by 
certain easy rules. Thus, if 



and if 



y=zAxn + 
dy 



dx 



= n K xn 



(1), 
(2), 




whatever kind of number n 
may be. 

11. In teaching beginners it is well to start on the assumption that 
students already possess the notions of the differential and integral 
calculus, and it is a teacher's duty to put before them the sjonbols 
used in the calculus at once. It is surely much better to do this 
than to evade the calculus in the fifty usual methods which we 
sometimes see adopted. Uniortunately many readers of this book 
are Hkely to be preparing for examinations in which only academic 
methods of elementary study are recognised, so we must keep our 
calculus symbols for the smaller print paragraphs. We may say, 
however, that we think even beginners in this subject will be able 
to understand the author's book on the calculus ; and if so, they 
will find the study of applied mechanics very gTcatly simplified. 
The language of the calculus is the natural, easy, simple language 
of the engineer ; but it is in tvriting, whereas most engineers only 
speak of rates and integTals. 

11 y is, known as a function of x, we can find — , which we 

dx 
may call u. Conversely, if w is known as a function of x, we can 



16 



APPLIED MECHAXICS. 



find y. The two symbols are -^. called a rate, or " the differential 
OX 

co-efficient of // with regard io jt" : and I m . <ir, eaUed '• the in- 
tegral of M with resrard to x." J 



12. The following f oimnlae will sxiffice for nearly all engineering 
calculations : — 



y 


dy 


i • 


■» 

V 


c 




! 


c 


C I 


Ar» 


n^x'^-^ 


Bi;» 


B , .. - 1 


A log. X 


A 

- i 


1 


Alog.s 


A^ 


mAtf^ 


A^ 


1a.- 

a 


Alog.{x-{-a) 


A 


A 


A log. Iz -r a) 


A sin. (or + &) 


« A cos. («: -f *) 


Aan. {ax^h) 


— — COS. \:fX — (/j 

a 


A COS. {ax -L b) 


— a Asin.(ffiE -f- *) 


Acos.(<u; + d) 


sin (<tc -I- o; 



and all ihe imporrant part of th: 
devoted to illtLtrating the use c i 



;'r- tiie ca.c-aiui is 




PSg.fiL 



B T 



13. For example, if z is the 
acsois-, 1 -i y is the ordinate 
PB c: :Lr oiiTe shown in Fig. 
; if the area between the 
curve and o x and p b, and any 
either ordinate nearer o. say d s, hie 
called A ; and if the area to q t be 
called A -f 5 A, o T being called 
a: + 5r, the area 5a = p a t b, and 
as Sr is made smaller and smaller 
this area gets more and more nearly 

(«/ -f - 5y) 8ar, as Q T is y -{- 5y. 

Thnfl 



APPLIED MECHANICS. 



17 



and as 5^ gets smaller and smaller tliis in the limit gets to be 

dA 

dx 
Hence, if we know ■{/ in terms of x, we can find a by integration. 
If we know a, we can find ?/ by differentiation. Thus, if we have 
tlie parabolic curve, 

i/ = ax^ .... (3) 

A= — ax^ -\- c . . . . (4), 
3 

a constant is added, because the rate of change of a constant 
is 0, and we wish our answer to be as general as possible. The 
value of c is fixed as soon as we settle from which ordinate a is to 
be calculated. Thus, if a is when x is 0, c is 0. If f {x) is the 
general integral of u, 

'^Xc 

u . dx 
x^ 

tells us to use x^ for x in the general integral, then to use x^ for x 
and subtract, or 
*x^ 

t . dx — ¥ [x^] - F {x^) .... (5). 



J* 



I 



If u is the ordinate of a curve, (5) means the area between the 
curve, the axis of x, and the two ordinates at x-^ and x^. 

Any summation which may be indicated by an area can be 
effected in this way by integration. It is very easy to recollect 



the rule that the rate of change of 
the integral of ^"'^ is ... . . x 



.m + 1 



most engineering problems 
for integTating ic ™ is 
in this particular case 



+ 1 
There is one case in which 
to us — namely, when m 



is nx'^ •■■, and that 
This rule suffices for 
the rule 



is - 1 



I- 



dx 



14. Integrations. — In a 

great number of problems we 
have the ordinates like ap, 
called the y, of every point 
of a curve, given us, the 
abscissa o a of the point 
being called x. 

If we are given a list of 
corresponding values of y 
and a;, we can draw the curve. 

Now, if A a represents to 
scale the area of a p r o, and 
if we find many points like q 




Fig. 6. 



18 



APPLIED MECHANICS. 



and join, them, we get a new CTure, which is said to represent the 

integral or area of the first. The areas may he found by means 
of a planimeter. I always us^ the following method myself. It is 
inaccurate to this extent, that the curve p. p is tak«^n to he a many- 
side«i polygi>n instead of a cuTre. But it lea»is to a very qniek 
solution of coraplieated-Lroking problems. Thus the following 
values of y are ijiven for the cr>rrespoTitiing values of x. 







r 


s 


y 











05 


1-869 




1 




1-869 


1-5 


1-756 




2 




3-625 


2-5 


1-657 




3 




0-2S2 


3-5 


1571 




4 




6-853 


4-5 


1497 




5 




8-3o0 


5 5 


1432 




6 




9-782 


6-0 


1-375 




7 




11157 


7-'5 


1-326 




8 




12-4S3 


8-5 


1-247 




9 




13-730 



It will he seen that we assume the or>iiiia:e at any p-ifnt }\\a 
x=Z b to be the average height of the curve from x = o to x=i 
Xotice that to get, say 6 853, we add 1-571 times 5j-, which is i 
to 5-282. TVe always take the values of y at equal intervals ia 
the values of x. If the above values of x had been 0, 0-05, 
010, 0-15, etc., hegioning with the first, we should have had to 
divide all the numbers in the last column bv 10. 



Exercise. — For values of :r=0-5, 1" 



., calculate variouj 



values of y iromy = 2 + 3 ar-fO 5 x^, and find the integrals. The 
true value of the integrals is x = 2 x -{- I'o or^-f |- x^. Calculate this 
for .r = 0, 1, 2, .... and note the errors in our numerical method- 
ic — ^The symbol -^ means the rate of change of -/ with rearard 
dx^ ^ dx 

d^u d^u 

-~ means the rate of change of -j^ with regard to x. 

It will he found that much of the difficulty which some 



to X 



APPLIED MECHANICS. 19 

students find in using tlie calculus is due to their not being able to 
differentiate f^ with regard to t, although they know how to 
differentiate x'^ with regard to x — the mere use of another letter 
than the one to which they have heen accustomed causing the 
difficulty. It is well, therefore, from the beginning to get used to 
many other letters, such as t, v, s, p, tv, etc. 

ds . , d'^s dv . 

If s is space and t time, -— is velocity v, and ^ = -^ is 

acceleration. Newton's symbols were: s for space (length), 

s for velocity, s for acceleration. 

d^s 
Example. — If acceleration -j^ = a, a. constant ; integrate with 

ds 
regard to t, and we have ^= velocity = «^ + b . . . . {6), where b 

is the constant which we always add to make the answer general. 
We note the meaning of b to be the velocity when ^ = ; and 

ds 
perhaps we had better use Vo instead of b, so that —=:at-\- vo 

Integrate again, and we have szzz^at'^ + Vot -\- e . . . . (7). Evi- 
dently c means the value of s when t=iO. (6) and (7) are well- 
known laws of uniformly accelerated motion. 

16. Example. — A chaiu of length x, hanging vertically down- 
wards, is being lowered from a capstan. Its weight per foot of its 
length is w. When the weight tvx is lowered through the distance 
Sx, the work done is ivx dx, and the whole work done by it from 
the time it is of no length, till it is of length I, is — 



I wx dx, or I wx^ , 



or, ^' 



or its total weight multiplied by half its vertical length. 

Suppose the chain to get thicker as more of it is let down ; say 
that wzzza + bx, then the above summation becomes — 

c^ r r^ 1 

\ {a + bx) X dx = \ {ax + bx'^) dx=. \ \ ax"^ -\- ^ b(x^ \ :=. 

^aP' + ^bl^. These two answers represent also the work done in 
lifting the chain. 

17. Example. — Fluid expanding through the volume ^v at the 
pressure p does work p. Iv, and 



^1 



is the work done in expanding from volume Vj to volume fg. Thus, 
suppose that expanding fluid follows the law 



pv^ = c or p = cv ^, 



20 APPLIED MECHANICS. 

where i and e are constants, we have to integrate ev"^ with 
regard to v. and according" to our rule the answer is 

'-L r - :. + 1 

Putting in the limits, our work is 

and this may he put in other shapes. Wlien the law m jw:=r, 
the work is e loa", — . 

18. The Compomid Interest Law. — If we are told that the rate 
of increase or dirninution of y with regard to x is propjrtional to y, 

saj that -- =r ay, then we know that j/ = A f <Lr, where A is 
any constant. 

19, The Harmonic Law. — If y = A sin. {ax -f b), we find that 
dy 
dx 
case: — 

If y = Acos. [ax + i), 

^ = - A <z sin- [ax -f J). 
dx 
Thns if y = A sin. <7a- ->- J), 

J- =^ K.a COS. (ff.r + *), 

^ = — A flS sin. {nx -f- 2>) = — a^. 

With this knowledge a student c-an put in a few words all the 
theory of Chapter XXY. He]wiIL see that simple harmonic motion 
is stated hy a- = « sin. [qt + e) Q). where a is the amplitude, 

q is 2 xf or — , where f is the firequency or t is the periodic 

T 

time, and e is the lead, a quantity introduced hecause x is not 
when f = : that is, we give generality to (1) hy assuming that 
we may hegin to count time from any position of the hcdy. 
Observe that the motion may he angular, x and a heing angles. 

— will then he angular velocity, and — -^^ will he angular 
dt ^ dt- 

acceleration. Differentiating twice we get 

velccity = '— =: aq cos. (j^t -\- e) . . . . (2), 

dt 

aceeXeration ■:= ^^ = — a^ sin. {qt ■\- e) . . . . (3). 
dt- 

Notice the siarn of — ^, and see how it aarees with the statements 
^ df^' 

fPx 
of Chapter XXY. If we only think of the numeric-al value of -j-^ 



APPLIED MECHANICS. 



21 



wc notice at once that x -h 



d^-x 



(4), and the square 



clfi q- 

root of this is T/2 7r, which is the rule fonnd in another way in 
Chapter XXV. 

20. Students will do well to graph on squared paper some 
curves like the following : — 

1. If y = 0-1 x^ + 3, take x = 0, x =s= 1, x = 2, etc., and in each 
case calculate t/. Plot the values of x and y as co-ordinates of points on 
squared paper, and draw the curve passing thi-ough these points. 

2. Graph i/ = a + bx. 

1st, when a^=z Q, b =: 1. 
2nd, when a = 1, ^ rr 1. 
3rd, when a; r= 1, 3 = 1-5. 
4th, when a = — 1, ^ := 1-5. 
5th, when a :=.!, b =^ — 1. 



Grraph y 
Graph y 



•1^. 
10 sin. 



5. Graph ?/ = 120 a; 

6. Graph y = 120 a: " 

7. Graph y = 120 a; " 

8. Graph y :^1Q x ^. 



(r-O- 



1-3^ 

0-7 



9. Graph y 

and also ?/ = 



3 y 25 - x^. 

21. The following observed numbers are known to follow a law like 
y — a -\- bx; but there are errors of observation. Find by the use of 
squared paper the most probable values of a and b. 



X 


2 


3 


44 


6 


7 


9 


12 


13 


y 


5-6 


6-85 


9-27 


11-65 


12-75 


16-32 


20-25 


22-33 



Ans., y — 2-b -\- 1*5 x. 

22. The following observed numbers are known to follow a law like 
y:= ax / (l + sx) : — 

Find by plotting the values of y/x and y on squared paper that these 
follow in a law y/x +sy — a, and so find the most probable values of a and s. 



X 


-5 


1 


2 


•3 


1-4 


2-5 


y 


•78 


•97 


1-22 


•55 


1-1 


1-24 



Ans., y = Sx I {1 + 2x). 



22 APPLIED MECHANICS. 

23. Vertical Line. — A line showing the direction in which 
that force which we call the resultant force of gravity acts. It 
is a line at right angles to the surface of still water or mercury. 

24. Level Surface. — A surface like that of a still lake, 
everywhere at the same level, and everywhere at right angles 
to the force of gravity or other volumetric force which is acting 
upon matter. It is not a plane surface. 

25. Curvature. — For any curve we can find at any place 
what circle will best coincide with the curve just there. The 
radius of this circle is called the radium of curvature at the 
place. But since we say, for instance, that a railway line 
curves much, when we mean that the radius is small, the 
name curvature is always given to the reciprocal of the radius. 
Thus, if the radius is 8 feet, we say that the curvature is -J. 
If at another place the curvature is ^., the change of curva- 
ture in going from the one place to the other is the diSerence 
between these two fractions. 

Curvature may also "be defined as the angle turned through by 
a tangent to the curve per unit of its length. A student ought to 
see for himself that the two definitions are the same. If in Fig. 5 
the distance along the curve between p and q is called ds, and if 
50 is the angle which the tangent at o. makes with the tangent at 
i», then the average curvature between p and q is bO/Ss. As p and 

Q become closer and closer, without limit, the curvature is -=-. 

as 
If a curve is defined bv its x and y co-ordinates, the curvature 

^^dx' ' \ ^^\dx) i • 

Example. — Find tbe curvature, where x = 0, of the parabola 

V =: ax"^. Here -— =z 2 ax and -^-^ = 2 a, so that the curvature 
^ dx dx? 

anywhere is 2 ^ 4- j 1 + 4 aH- 1 \ and at the ^'ertex where x = 
the curvature is 2 a. 

When, as in ordhiarv beams, -^ is small, it is evident that the 
' - dx 

d-y 
curvature mav be taken to be ^-^. This is what we take to be 

dx- 

true in the discussion of beams and struts. 

Exaviple. — In making 100 steps round a curve, luy compass, 
showing the direction of motion, changes from IS", to X.E. 
What is the average curvature? Answer — from N.to N.E. is 
45 degrees, or 07854 radians, and this divided by 100 steps, 
or '007854 radians per step, is the average cui'vature. The 



APPLIED MECHANICS. 23 

reciprocal of this, or 127*3 steps, is the radius of curvature, 
if the curvature is constant — that is, if the curve is an arc 
of a circle. 

Many unpractical rules will be found in books, requiring 
us to draw a tangent to a given curve at a given point, or to 
find its curvature there by trial. These are only academic 
suggestions. If half a dozen students get tracings of the same 
curve, and two points to measure the angle between the 
tangents there, they will obtain six very different answers. 

EXERCISES. 

1. Througli what angle must a rail 10 feet long be bent to fit a curve 
of half a mile radius? Ans., 0-22 degrees. 

2. Arc s is 10 feet long, radius r half a mile ; find the versed sine x — 
that is, the greatest deflection from straightness. Prove that, practically, 
«2 = 8 rx ; so that in this case x z= -0047 feet. 

/ X —X \ 

3. Find the radius of curvature of the catenary y = ^ ) ^ i^ >at 
the vertex. -^ ) \ 

Ans., a. 

4. Find the radius of curvature of the ellipse -^-\-j^ = l. (1) At the 
end of the major axis ; (2) at the end of the minor axis. 

Ans., (I) I, [2)1. 

26. Angle. — An angle can be drawn : First, if we know its 
magnitude in degrees ; a right angle has 90 degrees. Second, 
if we know its magnitude in radians ; a right angle contains 
1-5708 radians. Two right angles contain 3-1416 radians. 
One radian is equal to 57-2958 degrees. One radian has an 
arc, B c, equal in length to the radius A b or A c. It some- 
times gets the clumsy name " a unit 
of circular measure." Third, we can 
draw an angle if we know either its sine 
cosine, or tangent, etc. Draw any angle, 
BAG (Fig. 7). Take any point, p in 
A B, and draw p q at right angles to 
A c. Then measure p q, a p and A q 
in inches and decimals of an inch. 

p Q -^ A p is called the sine of the angle. A Q -^- a p is called 
the cosine of the angle, p Q H- a q is called the tangent of the 
angle. Calculate each of these for any angle we may draw, 
and measure with a, protractor the number of degrees in 




24 APPLIED MECHANICS. 

the angle. We shall find from a book of mathematical tables 
whether our three answers are exactly the sine, cosine, and 
tangent of the angle. This exercise will impress on our 
memory the meaning of the three terms. It will also 
impress upon us the fact that if we know the angle in 
degrees, we can find, by means of a book of tables, its sine, or 
cosine, or tangent ; and if we know any one of the sides A p, 
or P Q, or A Q, of the right-angled triangle A P Q and the angle 
A, we can find the other sides. 

Divide the number of degrees in an angle by 57*2958, and 
we find the number of radians. Suppose we know the number 
of radians in the angle bag, and we know the radius a B or 
A c, then the arc b c is 

A B X number of radians in the angle. 

Given, then, a radius to find the arc, or given an arc to 
find the radius, are very easy problems. 

A student becomes accustomed, on seeing an angle drawn 
on paper, to judge from a mere glance how many degrees the 
anole contains. It would be an advantage to acquire the 
habit of judging how many radians there are in the angle. 
What we mean is, that he ought to be as ready to think in 
radians as in degrees, and to do this he requires to be familiar 
with the size of a radian. 

EXERCISES. 

1. Draw an angle of 35°. Find ty measurement the sine, cosine, 
tangent, cotangent, secant, and cosecant of the angle, and compare with 
the numhers in a hook of tahles. Calculate the number of radians. 
Try if sin.^ 35° + cos.^ 35° z= 1 ; if sin. 35° -i- cos. 35° — tan. 35° ; if 
tan.'^ 35° + 1 = sec- 35° ; and if cot.^ 35° + 1 =: cosec- 35°. 

2. If a = 55°, iS = 20°, illustrate the following important formulae by 
numerical calculation : — 

Sin. (o + )8) = sin. a cos. )8 -\- cos. a sin. jS. 

Sin. (tt — /3) = sin. a cos. )3 — cos. a sin. )8. 

Cos. {a-{- &)-= COS. a cos. — sin. a sin. /8. 

Cos. (a — iS) = cos. a cos. /8 + sin. a sin. /3. 

Sin. a cos. ^ = 2 \ ^^^- (« + )8) + sin. (a — 3) } . 
Cos. a cos. /8 == 2 ^°^- (« + ^) + cos. (a — ^) I . 
Sin. a sin. Q=~y cos. (a — /3) — cos. (a + )8) j . 



Cos. 2a = 2 cos.2a — 1 — 1 - 2 sin 



APPLIED MECHANICS. 25 

3. Wliat are sin. 150°, cos. 130°, tan. 170°, cos. 240°, sin. 220°, tan. 
218°, sin. 290°, cos. 310°, tan. 320°? Express all these angles in 
radians. 

Ans., -5, - -6428, - -1763, - -5, - -6428, '7813, - -9397, '6428, 
- -8391; 2-6180, 2-2689, 2-9671, 4-1888, 3-8397, 3-8048, 5-0614, 5-4105, 
5-5851. 

4. Tlie sine of an angle is 0-25; find its cosine, tangent, cotangent, 
secant, and cosecant. Find the angle by actual drawing. How many- 
radians ? Afis., -9683, -2582, 3-875, 1-033, 4; 14°-5 ; -2528. 

5. What are the sine, tangent, and radians of 1^ degrees ? 

Ans., Each -0262. 

6. If in Fig. 7 a is 47°, and ap 5-23 feet, find a a and p q. 

Ans., AQ==3-567, Pa = 3-824. 

27. Angular Velocity. — If a wheel makes 90 turns per 
minute, this means that it makes 1*5 turns per second. But 
in making one turn any radial line moves through the angle of 
360 degrees, which is 6-2832 radians; so that 1-5 turns per 
second means 6*2832 x 1*5, or 9"4:248 radians per second. 
This is the common scientific way in which the angular 
velocity of a wheel is measured — so many radians per second. 
If a wheel makes 30 turns per minute, its angular velocity is 
3-1416 radians per second; n turns per minute mean 2irn 
radians per minute, or 27^?^ -i- 60 radians per second. One 
turn is the angular space traversed in one revolution. 

Exercise. — Show that the linear speed in feet per second of a point in 
a wheel is equal to the angular velocity of the wheel multiplied by the 
distance in feet of the point from the axis. 

28. Angular Acceleration. — The increase of angular 
velocity per second. If a wheel starts from rest, and has an 
angular velocity of 1 radian per second at the end of the first 
second, its average angular acceleration during this time is 
1 radian per second per second. 

EXERCISES. 

1. A shaft revolves at 800 revolutions per minute. What is its 
angular velocity in radians per second ? Ans., 83-79. 

2. A point is 3,000 miles from the earth's axis, and revolves once in 
23 hours 56 minutes 4 seconds. What is its velocity in miles per hour ? 

Ans., 787-5. 

3. The average radius of the rim of a fly-wheel is 10 feet. When the 
wheel makes 150 revolutions per minute what is the average velocity of 
the iim ? Ans., 157-1 feet per second. 

4. An acceleration of 1 turn per minute every second ; how much is 
this in radians per second per second ? Ans., -1047. 

5. A wheel is revolving at the rate of 90 turns a minute. What is 
Us angular velocity ? 



26 APPLIED MECHANICS. 

A point on the wheel is 6 feet from the axis ; what is its linear speed ? 
If its distance from the centre he increased hy 50 per cent., what does its 
speed hecome ? If at the same time the speed of the wheel increases 
50 per cent., what is now the linear speed of the point ? 

Ans., 9*42o radians per second; 56-55 feet per second; 84*82 feet per 
second ; 127 '2 feet per second. 

6. There is a lever, a o, 30 inches long, works ahout an axis at o. The 
lever is made to turn hy applying a force at a point k in a o, 15 inches 
from 0, so that b receives a velocity of 2 feet per second. What is the 
angular velocity of the lever ? 

If the same velocity had heen given to the point a instead of b, what 
would the angular velocity have been ? 

Afis., 16 radians per second ; 0*8 radian per second. 

29. Length of Belt. — Let d and d he the diameters of pulleys, c 
the distance hetween their centres, l the length of helt, let d -\- d 
he called s. Prove that for a crossed helt 

L = (I + e) s + 2c cos. e . . . . (1), 

where sin. = s -^ 2c. 

Example. — Find the length of a crossed "belt for pulleys of 
20 and 15 inches diameter, the distance of their centres apart being 
120 inches. First find 6, if sin. r= 35 -r- 240 — -1458. So that 
r= 8^ 23', or -1463 radians. Also cos. e = -9893. Hence we 
haA^e 

L = (1-5708 + -1463) 35 + 240 x -9893 = 297-5 iaches. 
Notice from the formula (1) that as l depends only on c and s. 
if c and s are the same, the same length of belt will do. Thus the 
same crossed belt does for any two corresponding steps in stepped 
cones if n + ^ is the same. It is exceedingly easy to calculate the 
sizes of these steps when we know the speeds. Thus the above- 
mentioned pulleys were 20 and 15 inches, and their speeds were as 
3 to 4. If there were two steps, and we want another pair to give 
a speed ratio, say, of 1 to' 2 using the same belt, we faiow that if 
their diameters are d and d 

D + (^ = 35, n = 2d. 

Hence, 3 ^ = 35 and d = llf , d = 231. 

Exercise. — The centres of two pulleys, 3| and 1^ feet in 
diameter respectively, are 10 feet apart. Find the length of 
crossed belt required."^ ^//5., 28 48 feet. 

If the belt is an open belt, let the student prove that the length 
L is 

L = -^ (D + <^) + (d - d) + 1c cos. 0, 

where sin. Q z=: [v> — d) -i- 2c. It is easy to show that the 
following answer, which is much easier to deal -s^-ith later, is 
practically coiTcct : 

L = |(D + ^ + 2.+ l-(J^\...(2). 
The length of belt depends upon d + d, c, and j) — d. 



29.'5-03". 



APPLIED MECHANICS. 27 

Uxercue. — Pulleys of 20 and 15 inches diameter, whose axes 
are 120 inches apart, are connected by an oxDen helt. Find its length. 

^»..,L = |(35) + 240(l+^^-^||5) = a 

The student ought to find the correct answer, and see how small 
'S the error in the use of our approximate formula. 

Suppose we have a pair of pulleys d^ and d-^, and we want 
another pair d^ and d<^ on the same shafts to work with the same 
length of belt, the ratio of d^ to ^2 being known. Putting the two 
sxpressions for l equal, we have 

D2 + 4 + 2V. ^'""-^ - '^'■^' = ""^ + ^1 + 2L ^""^ " "^''^ ■-•' ^^^- 
The right-hand side is known, and we know d^ in terms of d.2, so 
that it is easy to calculate. 

Exatnple. — In the above example of Dj = 20, f/j = 15, c = 120, 
let us calculate t>^ and d^, if D2 = 2 d^. Oui' equation (3) becomes 

^« + 2;r4i2o(^)^ = ^''^ + sr4T2oW^- 

The solution of this quadratic gives us fl'2 = 33 ; and, therefore, 
D2 = 66. In practice we usiially calculate d., and d^ as if the belt 
were crossed. These are nearly the right answers. Then, taking 
the D2 — (?2 so found, we find from (3) a corrected value of 
D2 -4" ^2> ^^^ "^® ^^^ this, knowing the ratio of d, to d^, to find 
them more correctly. We may, if we please, approximate more 
nearly by rej)eating this process, but this is seldom necessary. 

Example.— Otx the driving shaft, going at 100 revolutions per 
minute, the diameter of the first step is 20 inches. From this step 
the driven shaft, which is 120 inches away, is to go at 200 revolu- 
tions. From other steps the speed of the driven shaft is to be 150, 
100, 75, and 50 revolutions. Find the sizes of the steps if the belt 
is crossed. 

Here Dj = 20, d^ = 10, Dg + d.-^ =: 30, v^ = § d^. Hence | (fg + 4 
= 30, 2| ^2 = 30, d^=:12 inches, % = 18 inches. 

In the same way we find 1^3 =z 15, D3 = 15 ; d^ = n}, d^ = 12f ; 
^5 = 20,1)5^10. 

30. If a line, AB, makes an angle 6 with the horizontal, 
the projection of its length on the horizontal is A b cos. d. 
Its projection on a yertical line is A b sin. 0. 

Uxer else.— Draw two lines ox, oy at right angles to each other. 
Now draw lines o p, o q, o r, o s, of lengths 3, 5, 2 a, and 4 inches, ai^d 
making angles of 35", 72°, 130'', and 220° with ox." Find the projection 
of each line on o x and on o y, and the sum of the projections on o x 
and on o y. 

^y^*., 2-457; 1-545; -1-607; - 3-064 ; 1-721 ; 4-755 ; 1-915; - 2-571; 
-0-669; 5-820. 

If a plane area of A square inches is inclined at an angle 
with the horizontal, its area, as projected on the horizontal, 
is A COS. square inches. 



28 APPLIED MECHANICS. 

Try to prove that this must be so by dividmg the area into 
strips by horizontal lines. 

EXERCISES 

1. A plane area of 35 square feet is inclined at 20° to the hori- 
zontal : find its horizontal and vertical projections. 

Ans., 32*89 square feet ; 11-97 square feet. 

2. The cross-section (a cross-section always means a section by a 
plane at right angles to the axis or line of centres of sections) of a 
cylinder is a circle of 0*7 inch radius. Find the areas of sections which 
make angles of 25° and 45° with the cross-section. Note that the cross- 
section is a projection of any other section. 

Ans., 1-699, 2-177 square inches. 

3. The aljove cylinder is a tie bar of wrought iron. The total tensile 
load is 12,000 lb. ; how much is this per square inch of the cross-section? 
How much is it per square inch of either of the other sections ? 

^ Ans., 7794 lbs., 7063 lbs., 5512 lbs. 

4. The cross-section of a pipe is a circle of 15 inches diameter; what 
is the area in square feet ? If 1 3 gallons flow per second, what is the 
velocity vo ? What is the area of a section at 28° to the cross-section ? 
What is the velocity v, normal to this section, if normal velocity x area = 
cubic feet per second ? Show that v is the resolved part of Vo normal to 
the section. Ans., 1-228, 1-7 feet per second; 1-39, 1-5 feet per second. 

5. Part of a roof, shown in plan as 4,000 square feet, is inclined at 
24° to the horizontal; what is its area? Ans., 4378-7 square feet. 

6. A tie bar or short strut of 2 square inches cross-section ; what is 
the area of a section making 45° with the cross-section? If the total 
tensile or compressive load is 20,000 lbs , how much is this per square 
inch on each of the sections? Resolve the total load normal to and 
tangential to the oblique section, and find how much it is per square inch 
each way. Ans., 2-828 square inches; 10,000 lbs., 7,070 lbs., 5,000 lbs. 



29 



CHAPTER II. 

VECTORS. RELATIVE MOTION. 

31. Any quantity which is directive is called a vector quan- 
tity — for example, a velocity or a force. It can be represented 
by a line. Its amount can be represented to some scale by the 
length of the line. The clinure of the line and an arrow-head 
represent the clinure and sense of the vector. Yector quanti- 
ties are distinguished from scalar quantities, such as a sum of 
money, the mass of a body, energy, temperature, etc. 

The resolved part of a vector in any new direction is 
represented by the projection of 
its representative length in the new 
direction. Thus in Fig. 8, if op 
represents to scale a velocity or a 
force, its resolved part in the direc- 
tion X is o A, the amount of which 
is o P cos. A o p, and its resolved part 
in the direction o y is o b, the 
amount of which is p cos. bop. 
Thus, if a ship is going at 9 knots north-eastward, the northerly 
component of its velocity is 9 cos. 45°, or 6*363 knots, and its 
easterly component is 6-363 knots. 

If a body has an acceleration of 20 feet per second per 
second in the direction 25° east of north, the northerly com- 
ponent of this is 20 COS. 25°, or 18*13 feet per second per 
second; and the easterly component is 20 sin. 25°, or 8*452 
feet per second per second. 

If a force of 30 lbs. is in a northerly direction, its com- 
ponent in a north-easterly direction is 30 cos. 45°, or 21*21 lbs. 

32. The resultant of two or more forces is a force which 
might be substituted for them without changing the effect. Jf 
two strings pull a small body with forces of 5 lbs. and 7 lbs. 
(Fig. 9), and if the angle between p ^ 
them is 30°, draw o P equal in 
length to 5 inches, and make the 
angle qop equal to 30°. Make 
the length of o q 7 inches. Com- Fig. 9. 

plete the parallelogram q p R, 

and draw the diagonal o r. Measure r in inches ; we find 

it to be 11*6 inches, so that the resultant of the two forces 




30 APPLIED MECHANICS. 

is 11 "6 lbs. One string acting in the direction or mth a 

pull of 11-6 lbs. will produce the same effect at o as the two 

strings did. In using this construction, take care that the 

arrow-heads are confluent — that is, that they all point away 

from 0, or they all point towards 0. Suppose when the two 

strings were acting we had found by experiment that a third 

D string E (Fig. 10) would just prevent the 

-^ two strings from causing motion at o, then 

L,^^r--i~-^ — ^ experiment would also show that the force 

Fig. 10. ii' E, which may be called the equilibrant 

of o P and Q, is exactly equal and opposite 

to the resultant of o P and o q 

We see that when o Q and o p are given, and the angle 
between them, we may use the above principle, called the 
parallelogram of forces; or, what comes to the same thing, the 
triangle of forces. Draw o q one of the forces, draw q r the 
other, and let their ari'ow-heads be circuital ; then the non- 
circuital force R is the resultant ; or a circuital force r o 
would be the equilibrant. It is in this way that we find the 
resultant or vector stim of any two vectors. 

In vector language, oq + qr = or, or oq = or-qr. 
This principle is very easy to express ; to be able to apply it 
implies a considerable experience. TS^e mention it now merely 
to introduce a few exercises. 

It is veiy easy to solve problems graphically. The student 
must work some numerical exercises, and test his answers 
graphically. 

Example. — Two forces, op and o q, of 5 lbs. and 7 lbs. 
respectively, act a point, at an angle of 56° ; find their 
resultant by calculation. 

In Fig. 11 OP and OQ represent, to scale, the two forces, 
the angle p o Q being 56°. Find the resolved part of o p 
(Art. 31) in the direction o x. say ; it is o a = o p cos. 56° = 
5 X -5592 = 2-796. Xow obtain the resolved part of o p 
in the direction o y, taken at right angles to ox; it is 
B = P . COS. p B = 5 sin. 56° = 5 x "829 = 4-145. In- 
stead of the given forces we now have O a and o Q acting 
along ox, and OB along o Y. Draw oh (Fig. 11) equal to 
o A -h OQ, that is, 9 796, and ov = 4'145. The resultant of 
these is o r, and we have 

or2 = OH-^ -h ov^^ == (9-796)^ + (4-145)2 = 24-996, 

. • . E r= 5 very nearly 



APPLIED MECHANICS. 



31 



To obtain the direction of o r, we have 



tan. a = — = 



RH 4-145 



OH 9-796 

. a = 23° nearly, 



= -4231 ; 



The student should test this result by finding o R graphi- 
cally by the method explained above. He should also observe 
that if the angle p o Q had been, say, 130°, OA would have been 
directly opposed to o q, in which case o h would have been 
obtained by subtracting o a from o Q. It is of the utmost 
importance that he should work many exercises similar to 
this one, so as to become familiar with the method. 

It will be shown later (Art. 94) that the same method 
is employed for calculating the resultant of any number of 




forces. In every case we resolve the forces along any line o x, 
and let o h represent their resultant ; then resolve along o y 
taken at right angles to o x, and let ov represent the resultant 
of these resolved parts, o r and a are then easily calculated. 

EXERCISES. 

1. Force of 20 lbs. at an angle of 72° with the horizontal; what are 
its horizontal and vertical components? Ans., 6-18 lbs., 19-02 lbs. 

2. A man walks towards the north-north-east at 4 miles per hour ; at 
what rate is he getting towards the east ? And at what rate towards the 
north? Ans., r53 miles an hoiu' ; 3-69 miles an hour. 

3. Forces o p of 10 lbs. and o q of 7 lbs. at an angle q o p of 35° ; 
find the resultant. Test your answer by working the problem graph- 
ically. Ans., 16-24 lbs. inclined 14° 18' with the force 10 lbs. 

4. On a horizontal surface there is a normal pressure of 4 tons per 
square inch and a tangential force of 3 tons per square inch from N.E. to 
S.W. What is the total force per square inch ? 

Am., An oblique pressure of 5 tons per square inch, making an angle 
tan ~ ^f with the N.E. to S.W. du-ection in a vertical plane. 



32 APPLIED MECHANICS. 

5. An ariAdl is carried by three ropes, which make angles 20°, 30°, 25° 
with the vertical; the tensions in the ropes are known to be 1,000 lbs., 
700 lbs., and 1,200 lbs. What is ihe weight of the anvil ? If the hori- 
zontal components of these three forces are drawn as balancing one 
another, find the azimuthal angles which the vertical planes through the 
ropes make with each other — that is, find the angles in the plan of the ropes. 
Ans., 2,633 lbs. ; 85° 47', 137° 43', 136° 30'. 
33. If the magnitude of o p (^ig. 11 ) is called p, of o q is a, 
of o R is R, and if the angle q o p is 0, it can be proved, that 

R = A/p2 + 0,2 _|_ 2 p Q, cos. d. 
It will be seen from Fig. 1 1 that 

, RH OB p sin. 

OH OA+OQ, p. COS. 6 + Q 

If is a right angle, p and o, are the resolved parts of r in 
their two directions — 

R = a/ p2 + q2^ a,nd tan. r o a = p/a. 

EXERCISES. 

1. P = 3, a = 4, = 90". Then r = 5 and a is an angle whose 
tangent is 0*75. [The neat way of making this statement is to write 
a = tan. -1 0-75.] 

2. p = 3-045, Q == 7-462, = 37°; find r, a. Ans., r = 1006. 

a=10°30'. 

3. p = 3-045, Q, = 7-462, = 143° ; find r, a. Ans., r = 5-353. 

a = 20°. 

4. p = 12-06, a = 1002, e = 184° ; find r, a. Ans., r = 13-05. 

a = 187° 42'. 

34. A river flows at 1 mile per hour ; a swimmer has a 

velocity of 2 miles per hour relatively to the water. What 

is his velocity relatively to tlie bank *? Ihis will depend upon 

the direction in which he swims. 

Make ab represent 1 mile per hour down the river (as 
some students cannot get out of their heads the wrong notion 
that these lines represent distance, imagine 
the drawing to be infinitely small but 
*-^^ greatly magnified merely that it may be 

examined) ; make b c represent the velocity 
of swimming to scale, the direction and 
sense being correct. Take care that the 
Fig. 12. arrow-heads are circuital. Then a c is 

the sum of the two velocities possessed 
by the swimmer, and is therefore his velocity relatively to the 
bank. Let the student draw b c in all sorts of directions 
and reflect upon his answers. If he has ever swam in a broad 
river or has watched a swimming dog trying to reach his 
master, he will understand his answers more readily. 



APPLIED JIECHANICS. O.S 

EXEEOISES. 

1. Given the above velocities and that the stream flows clue south, if 
the absolute motion of the swimmer is to he south-east, in what direction 
ought he to swim? Ans., 24" 18' S. of E. 

2. A steamer moves westward at 10 feet per second ; a hoy throws a 
hall across the deck northwards at 4 feet -per second. What is the velocity 
of the ball relatively to the water ? 

Ans., 10-77 feet per second, 21*^ 48' N. of W. 

3. A steamer has a velocity of 14 knots due west ; the wind blows with 
a velocity of 7 knots from the north. What will be the apparent velocity 
of the wind to a person on board the steamer ? 

Ans., 10-7 knots from W. 26° 34' N. 

4. Velocity of a ship westwards 10 feet per second ; velocity of a ball 
on deck 5 feet per second north-east relatively to the ship. WTiat is the 
total velocity of the ball ? Ans., 7-37 feet per second, 28° 40' N. of W. 

5. If the total velocity of the ball is 12 feet per second northwards, 
what is its velocity relatively to the ship ? 

Atis., 15-62 feet per second, 50° 12' N. of E. 

6. A railway train is going at 30 feet per second; how must a man 
thi'ow a stone from the window so that it shall leave the train laterally at 
1 foot per second, but have no velocity in the direction of the train's 
motion ? 

Ans., 30-02 feet per second ; at an angle of 178° 6' with direction of 
motion of train. 

35. A bicyclist is ordered to travel so that he shall be more to 
the north at the rate of 3 miles every hour, and he must keep to 
roads. Notice that if he is on a north and south road his task is 
very easy. If his road is directed N.AV., he must travel at 4-242 
miles per hour. If his road is W.N.W., he must travel at 7-839 
miles per hour. If his road is due west, his task is an impossible 
one. If his road makes an angle 6 with due north, he must tl•a^•el 
at the rate of 3 -^ cos. 6 miles per hoiu\ 

36. Water flowing from the inner to the outer part of a 
motionless wheel of a centrifugal pump is guided by vanes to 
follow a curved path. SujDpose its radial velocity v,- known ; show 
that its real velocity anywhere is v,- ~ cos. 9, if is the angle 
which the vane there makes with the radial direction. 

A student will find this an excellent graphical exercise. Draw 
circles of 1 and 2 feet radii to represent the inner and outer 

cylindric sm-faces of the wheel of a a 

pump. Draw any shape of vane connect- 
ing these, but you had better take a 
shape fi'om an actual wheel (see Art. 
427). Let the angle at a with the tangent 
to the circle there (Fig. 13) be 18|^*^. 
Now imagine a particle of water travell- 
ing out radially at 0-1 foot per second. ^ 
Imagine it to take 10 seconds to get to q. '°' ' 
Mark its successive positions along the vane. Vou had better 
also trace its positions backward for a few" seconds towards a from s. 
Now imagine the wheel to revolve about its centre so that a tixoves 




34 APPLIED MECHANICS. 

at 0-3 foot per second. Map out the real positions of the points 
which you found on the vane, and therefore the real path of a 
particle of water. Try to follow it after it has left q, assuming 
plenty of space outside ; hut this is a prohlem which you perhaps 
may return to later, 

37. Let p (Fig. 14) he a point on the rim of the wheel of a 

centrifugal pump. Suppose that we know the velocity of r, and 

represent it by the distance p b ; also that we know the velocity 

of the water along the vane of the wheel at p, and represent it 

hy the distance a p. Now, a particle of 

-p water at p has both these velocities ; a p 

"^ ^^^^^^'" — 1 — ■"^^-^ relatively to the wheel, together with pb 

y^ \ ^s because the wheel is in motion. Hence the 

, \ total velocity of the particle of water is repre- 

^ I \ sented in direction, amoimt, and sense by the 

_. . vector sum a p -\- p b, and we can either use 

the parallelogram method or the triangle 

method to find it. 

Exercise. — The rim of the wheel of a centrifugal pump goes at 30 feet 
per second; water flows ladiaHy at 5 feet per second; the vanes are 
inclined backward at an augle of 35° to the rim. What is the absolute 
velocity of the water? What is the component of this parallel to the 
rim? Ans., 23-4 feet per second; 22-8 feet per second. 

AATien we know the velocity of water before it enters a 

wheel and the velocity of the wheel at the place, and we wish 

the water to enter without shock, this simply means that the 

total velocity of the water the instant after it enters the wheel 

shall be exactly the same as before it entered. Thus in Fig. 15, 

if the velocity, c p, of water is known before it enters the turbine 

wheel, and v b represents the velocity of the wheel, find the 

velocity which, added to p b, will have c p for a resultant. It is 

p a if pa represents the velocity c p to scale. Make, then, p a the 

direction of the vane at p. The water will flow in the direction p a 

^ relatively to the wheel, and it has also the 

^\ velocity p b because it moves with the wheel ; 

""^^ its total velocity is the same as before it 

^^_JiJ^_^^^^^ entered, and it has entered without shock. 

^^^^ /^ — ~— — ^ We don't much care how the vane curves 

/ "^\\ / afterwards so long as it curves gTadually; 

/' "^v / it is its direction at p that is important. 

AT"— -— ~^.J:^; ■ ^ 

! "Q Exercise. — The inner circumference of a 

; centrifugal pumx) wheel goes at 15 feet per 

* p.^ ^ second ; water approaches it radially at 5 feet 

°' ' per second. W^at is the angle of the vanes 

if the water is to enter without shock? Ans., 18° 26'. 

Note that the angle chosen by us for the vane at a (Fig. 13) 
enabled the water to enter the wheel without shock. We may at 
once say that it is only the angles at a and at q that are of real 



APPLIED MECHANICS. 35 

importance in the design of a pump. Th.e actual shape of the vane 
is unimportant so long as the curve is fairly direct. But the 
exercise ought to be worked carefully throughout. 

We made the tangent of the angle at a equal to the radial 
velocity -1, divided hy the velocity of the wheel at a (see Art. 33), 
We always endeavour to have this sort of relation true at the inner 
side of the wheel of a centrifugal pump or turhine. 

38. Usually we consider the earfcli and frame of a machine 
to be fixed. The student will find it very instructive to think 
of some link or wheel as fixed (which he usually thinks of as 
moving) and now note what the motions are. One simple 
example of this is given in Art. 467, and we there show that 
to understand the relative motions in a four-link mechanism 
is to understand the motions in a very great number of other 
mechanisms. 

Train a is passing train b. Looked at by an observer on 
the ground, how very different is their appearance from what 
it is to an observer in either train ! 

The mathematics of the subject is quite easy. It is simply 
this : If a, h, c, etc., are lines drawn representing what may be 
called the absolute displacements of points a. b, c, etc., or the 
rotations of bars about axes, then their displacements relatively 
to a frame F, whose own absolute displacement or rotation is 
/, are a — f,h — /, c — /, etc., the — sign meaning vector 
subtraction. 

But mathematics does not satisfy us ; we want the instinct 
of comprehending easily these relative motions. That we do 
not possess it is evidenced by the fact that all the mechanisms 
of Art. 467 do really seem to us different, and that we need to 
give the name epicyclic train to a train of wheels when the 
framework which connects their centres is allowed to move 
instead of remaining at rest, as it does in our usual way of 
studying things. Notice that this, which ought to be an easy 
subject, follows in smaller printing. 

39. Thus, for example, in Fig. 16 we have a train of three wheels 
(the student ought to take two or four or 
more). When the frame f is at rest (that 
is, we take sjDeeds relatively to the frame) 
let A, «, c has'-e the angular velocities a, 
ha, ca (evidently in the figui^e h is negative). 
Now let r have an absolute angular velo- 
city, /, and the absolute velocities of the ^^S'- ^^' 
wheels are a + f, ba -\- f, ca ■{■ f. 




36 



APPLIED MECHANICS. 



1. Suppose A is at rest ; a +/= 0, a i= — /. b's velocity is 
{— b -\- l)f; c's velocity is (- c -f 1)/. Thus, let c have the 
same number of teeth as a ; so that c = 1, c's velocity is 0. 
If c has 100 teeth, and alongside it on what is practically the same 
spindle, let there be wheels of 99 and 101 teeth also gearing with b. 
Evidently the absolute speeds of the three will be (since c has the 

,, 1 100 100 . 100\ 1,^,1 „^ 

three values , , and j as — — to to . \Mien 

99' 100' 101/ 99 101 

the arm goes round and the motions of the three wheels are 
observed, we call this Ferguson's paradox. If we conld, like flies, 
move with the arm, there would be nothing paradoxical about it. 
Notice that a, b, and c may be bevel wheels, as shown in Fig. 17. 

2, Simpler Case. — Wheels a and b connected by arm f ; aim 




rotates. The absolute rotational velocity of b is 0. What is a's 
velocity ? 

Here ba -{-/=: 0. Hence a=z — fjb ; so that a's velocity is 

_ fjh -\^ f or/A - iy Thus, let b = - 1, as it is in 

Watts' sun and planet motion ; a's velocity is 2/. In this 
case, however, because of the angularity of the connecting- 
rod, B has some angular velocity, fluctuating between, 
say, -\- P and — ^. Hence (since ^» = — 1), — o +/= + 3, 

and a's A^elocity is 2/ + fi. 

3. In Fig. 17 we have a, n, and c, three bevel wheels, b may 
be carried round the axes of the others on a spur wheel d. Suppose 
we are looking from the point D at every wheel, and the numbers 
of teeth on a and c are as 1 to c, then relatively to d the speeds of 
A and c are a and — w. If d rotates at speed /, the absolute 
velocities of a and c are a -\- f and - ac + f. 

Example. — Let c z=z I ; then, if a's speed a + / is called a, 
so that a = a — f, c's speed is — o + / or — a -\- 2/. Thus, 



APPLIED MECHANICS. 



37 



suppose A 

is 



following 
tatle : — 



goes at 20 revolutions per minute or a = 20, then c's 
— 20 + 2/; so that if / is gxadually changed in the 
way, we get the following speeds for c shown in the 

Again, we may imagine the 
speed of p to keep constant and 
that of A to vary, and we get the 
same result. 

By means of a cam we may 
gTadually change from negative 
to positi-s'e velocities, but here we 
have a very much better means 
by ordinary gearing. 

In Fig. 17, if the shaft of a is 
rotated by coned pulleys, by which 
it may be given varying speeds, 
and if a is kept rotating at a fixed 
speed, c gets speeds which may be 
negative or positive or zero. 

Exercise. — An epicyclic gear consists of an annular wheel a of 72 teeth, 
a pinion b, and a spur wheel c of 40 teeth concentric with a. The arm 
which carries the axis of b makes 30 revolutions per minute. (1) If a 
be a dead wheel, find the revolutions of c. (2) If c be a dead wheel, find 
the revolutions of a. Ans., (1) 84 ; (2) 46f. 

Again, if a makes 4 revolutions and c 6 revolutions in the same 
direction, find the revolutions of the arm. Ans. , 4|. 

Exercise. — In a horse-gear for driving a chaff-cutter, the bracket 
that holds the pole supports also a short horizontal shaft carrying a bevel 
wheel of 31 teeth and a bevel pitjion of 16 teeth. The pinion gears into 
a horizontal bevel ring of 80 teeth that is stationary, and forms part of 
the framing. The bevel wheel of 31 teeth also gears with a bevel pinion 
of 22 teeth which is loose on the central vertical axis, and this pinion 
carries with it a bevel wheel of 60 teeth that gears with a pinion of 
16 teeth on the high speed horizontal shaft. Find the number of revolu- 
tions of the high speed shaft for each circuit of the horse. 



Speed of F. 


Speed of 0. 


4 


- 12 


6 


- 8 


8 


- 4 


10 





12 


4 


14 


8 


16 


12 


18 


16 


20 


20 



38 
CHAPTER III. 

WORK AND ENERGY. 

40. Work. — To do work it is necessary to exert a force 
through a certain distance in the direction of the force. Thus, if 
we exert a force of 20 lbs. through a distance of 6 feet, we do 
20 X 6, or 120 foot-pounds of work. If a body of 5 lbs. weight 
changes its level by the amount of 10 feet, whether it does this 
by a direct vertical fall or rise, or is moved up or down an 
inclined plane or curved surface, so long as there is no friction, 
the amount of work given out by the body in falling or given 
to it to make it rise is always the same, 5 x 10, or 50 foot- 
pounds. 

Example. — The weight in si certain clock is 20 lbs., and 
after being wound up it can fall through a distance of 40 feet. 
Suppose we wish to alter this height, making it 10 feet ; what 
weight must we use ? Evidently the work given out by the 
new weight in falling 10 feet must be equal to that given out by 
the old weight, or 800 foot-pounds. In fact, the new weight 
must be 80 lbs. Of course we must apply this weight to the 
clock by means of a block and pulleys, or we must reduce the 
diameter of the drum proportionately ; and if in applying it we 
introduce more friction than there used to be in the clock, we 
must further increase the weight, so as to be able to overcome 
this friction. 

The work done by a force is well illustrated by the pulling 
of a tra,incar. If the pulling force p lbs. is not directly along 
the track, but makes an angle Q with it, the effective force, or 
the resolved part of p in the direction of motion is P cos. Q, 
and this, multiplied by the distance moved through in feet, is 
the work done in foot-pounds. 

When a body is pulled up a curve the work done in over- 
coming the force of gravity (we are neglecting work spent in 
overcoming friction) is simply the weight of the body multiplied 
by the difference in level. 

Thus in Fig. 18 the work done in moving- a body of weight w 
from A to B along the curve is simply w y, where y is the difference 
in level of a and b. 

Proof : — Let the co-ordinates of any point p be x, y ; and of q, 9 
point indefinitely near to p, a; -j- hx, y + hy. 



APPLIED MECHANICS. 



The weight w resolved in the direction of the tangent at p is w 
sin. e, and this multiplied by the distance p a (we are supposing 
that the tangent at q is parallel to the tangent at p, as a is 
supposed to be indefinitely near to p), or p a . w sin. is the work 
done against gra^dty in pulling the body from p to q. Therefore 
the whole work done in puEing the body from A to b is the sum 




A N 

Fig. 18. 
of all such terms as p a . w sin. 9. But' p Qsin. 6 is qr, or 
in the limit dy. Therefore the whole work done is 

wr/y or wy. 





or 



i; 



Gravity does this work when the body falls. But, both 
when a body is moved up and down, energy is wasted or con- 
verted into heat in overcoming friction. Hence, when a 
weight, w lbs. , is lifted in level h feet, the useful work done is 
wh, but there is energy wasted. Again, when w falls, 
gravity does the work wh, but there is energy wasted. If 
we are depending upon the total store w/i to drive machinery, 
the \iseful work done is less than wA, the difference being 
wasted, or rather converted into heat, by friction. 

41. Horse-power. — One horse-power is the work of 
33,000 foot-pounds done in one minute. Power means not 
merely work, but work done in a certain time ; the time 
rate of doing work. The work done in one minute by any 
agent divided by 33,000 is the horse-power of that agent. 
In a steam-engine the piston travels four times the length 
of the crank in one revolution, and all this time it is 
being acted upon by the pressure of steam. If the mean or 
average pressure urging the piston is 60 lbs. per square inch, 
and the area of the piston is 150 square inches, then the total 
average force urging the piston is 150 x 60, or 9,000 lbs. If the 
crankj whose length is 0*9 foot, makes 70 revolutions per 



40 APPLIED MECHANICS. 

minute, then the piston travels 4 times 0*9 x 70, or 252 feet 
per niinnte, so that the work done in one minute is 9,000 x 252, 
or 2,268,000 foot-pounds. Dividing this by 33,000, we find 
the horse-power of the steam-engine to be 68-7. The mean 
pressui'e is best found by the use of an indicator which draws 
for lis an indicator diagram. Measuring the pressures at ten 
equidistant places on this diagi-am, adding them together, and 
dividing by ten, gives the average pressure. Or, measure the 
area by means of a planimeter in square inches ; divide by the 
extreme length of the diagi-am parallel to the atmospheric line; 
tiiis gives the average breadth, and therefore the average 
pressure to scale. 

As the pressure of st^am is usually given per square inch, 
it is usual to take the diameter of the cylinder in inches, but 
distances passed through by the piston are evidently to be 
measured in feet. 

Example. — We find by a spring balance that some horses 
or a steam-engine have been pulling a carriage with an average 
pull of 120 lbs. during one minute, the space passed over in 
the minute being 500 feet ; what is the horse-power expended 
on the carriage 1 Here 120 lbs. act through the distance of 
500 feet, and the work done in one minute is e^vidently 500 x 
120, or 60,000. Dividing by 33,000. we find the horse-power to 
be 1-818. 

42. Energy is the capability of doing work. When a weight 
is able to fall, it possesses potential energy equal to the weight 
in lbs. multiplied by the change of level in feet through which 
it can fall. When a body is in motion, it possesses kinetic 
energy equal to half its mass (its weight in London in pounds 
divided by 32-2 is its inertia, which is usually, but we think 
unwisely, called its mass), multiplied by the square of its 
velocity in feet per second. (See Art. 190.) 

Exam-pie. — A body of 60 lbs. is 100 feet above the ground, 
and has a velocity of 150 feet per second; what is its total 
amount of mechanical energy ? — that is, what energ}' can it 
give out before it reaches the ground, and becomes motionless? 
Here the potential energy is 60x100, or 6,000 foot-poimds. 
Its kinetic energy is 60x150x150^-64-4, or 20,963 foot- 
pounds. So that the total amount is 26,963 foot-poimds. 

Suppose this body to lose no energy through friction with 
the air, and suppose that, after a time, it is at a height of 20 feet 
above tlie ground ; find its velocity. Answer : Its potential 



APPLIED MECHANICS. 41 

energy is now 60x20, or 1,200 foot-poimcis, therefore its 
kinetic energy must be 25,763 ; and evidently this, multiplied 
by 64'4: and divided by 60, gives 27,652'29, the square of the 
new velocity. Its velocity is therefore 166-3 feet per second. 
In such a question we are not concerned with the direction in 
which the body is moving. It may be a cannon-ball, or a falling 
or rising stone, or the bob of a pendulum. Given its velocity 
and height at any instant, we can find for any other height 
what its velocity must be, or for any other velocity what its 
height must be. 

On a Switchback railway, the loss of energy (roughly pro- 
portional to distance travelled) every journey is represented 
by the lift of a few feet which is effected by the attendant 
at each end before making a fresh start. Neglecting this loss, 
it is easy to calculate the velocity at any place if we know the 
vertical depth of the place below the starting point. The 
deepest places on the line are places of greatest velocity ; 
the highest places are those of least velocity. In a bob of a 
pendulum we see a continual conversion of kinetic into 
potential and of potential into kinetic energy, the total store 
remaining constant, except in so far as friction is converting 
mechanical energy into heat. (See Art. 192.) 

The energy of a strained body — for example, a strained 
spring — is another store of mechanical energy. It is an excellent 
laboratory experiment to measure how much energy a spring 
can store without getting permanently hurt in shape or broken. 
This store of energy is called its resilience. As the elongation 
of a spring is proportional to the load, if we gradually increase 
the load from to w lbs., the elongation increases from to 
X feet, so that the average force for the whole length being 
^ w, the energy stored is | w rr foot-pounds. 

A weight vibrating vertically at the end of a spiral spring 
gives a good example of the continual conversion of the three 
kinds of mechanical energy into one another. The total store 
gradually diminishes, partly by friction in the atmosphere, but 
greatly also by an internal frictional resistance or viscosity in 
the material of the spring. The student will find it interesting 
to compare the behaviour of a spring of steel and a spring of 
indiarubber in this respect ; the viscosity of the indiarubber 
damps the vibrations with great rapidity. Many hours may 
be well spent in studying these phenomena, making quantita- 
tive measurements. 



42 APPLIED MECHANICS. 

Air and other fluids in a compressed condition contain 
stores of energy. 

43. A student may employ his leisure in calculating the 
possible stores of energy in 1 lb. of various materials : 1 lb. of 
hydrogen, 4-8 x 10" foot-pounds ; kerosene, 2 x lO'' foot-pounds; 
coal, 12x10^ foot-pounds (the weight of oxygen or air for 
combustioQ is not counted in) ; 1 lb. of cast iron in rim of 
pulley at highest speed to produce greatest working stress, 
1,000 foot-pounds ; steel spring of the best kind, 270 foot- 
pounds ; usual spiral spring of round wire, 135 foot-pounds. 

The heat given to 1 lb. of water to raise it from 0" to P C. 
or from 99° to 100° C. in temperature, is nearly the same. 
This is a unit of heat energy. Joule showed that this is 
equivalent to 1,400 foot-pounds.* When energy is in the heat 
form, a heat-engine may be used to convert part of it intoi 
mechanical energy ; unfortunately the amount convertible,, 
depends upon how high is the temperature of the stuff contain- 
ing the heat energy above the temperature of the refrigerator 
or exhaust. Hence it is that in the very best steam-engines we 
seldom convert more than one-seventh of the heat energy of 
the steam into mechanical energy, the other six-sevenths being 
degraded in temperature, and going to the refrigerator useless 
for our purposes. In discussing heat-engines we express ali 
energy in foot-pounds. 

The energy store is most intense as to volume (we do not 
include the weight of the air needed for combustion) in a pound 
of kerosene, being about half as much again as in a pound of 
coal. We look forward to the time when no heat-engine 
will be needed in the conversion, and we may then be able to 
convert over 90 per cent, of the energy of a pound of kerosene 
into the mechanical form, as at the present time the chemical 
energy of zinc is convertible in a battery or electric motor, or 
the chemical energy of oats and other food is convertible in the 
animal machine, which is probably a gas battery and electric 
motor. 

44. When the engine of the Finsbury College is working 
mainly for the electric light and possibly one electric motor, the 
students have sometimes, during a long-continued measurement, 
been able to trace what becomes of the energy of one pound 

* The latest determination for the mean specific heat of iivater from 0° C. 
to 100*^ C. is 1399 foot-pounds, or for 1 gramme it is 0'99o calorie ; 1 calorie, 
the heat to raise 1 gramme of water from 10° C. to 11*-^ C. is 4 '2 .Joules or 
4-2 X 10" ergs. The heat from 20^' C. to 21* C is l-30th of 1 per cent. less. 



APPLIED MECHANICS. 43 

of coal. Tliey had previously measured the chemical energy 
in a pound of coal, and tested all sorts of instruments and 
ideas used in their measurements. 

The energy of 1 lb. of coal is, say, 12,000,000 foot-pounds; 
of this 4,000,000 go up the chimney or get wasted by radiation 
as heat, and 8,000,000 foot pounds of heat energy reach the 
engine in steam ; of this only one-thirteenth, or 600,000 foot- 
pounds is converted into mechanical energy and given to the 
piston; the other twelve- thirteenths go ofi to the condenser and 
are wasted. Only 500,000 are given out by the engine to the 
long shaft which drives the dynamo machine, about 400,000 
foot-pounds leave the dynamo as electric energy, and part is 
wasted by conversion into heat in the conductors to the lamps. 
At the lamps we have about 370,000 foot-pounds of electric 
energy converted into heat and light. If any of the electric 
enei'gy or all of it is given to an electric motor, perhaps more 
than 90 per cent, of it will be converted into mechanical 
power. 

Men who make measurements of this kind get to have very 
clear ideas as to the various forms of energy and the fact that 
it is indestructible, and cannot be wasted, although it may 
alter into forms in which it may not be available for use ; and 
therefore we say that it is wasted. 

The very best steam-engines use more than IJ lbs. of coal 
per hour for each horse-power given out. We cannot hope for 
much improvement; this is about one useful for nine total. 
Gas-engines using Dowson gas already give out one horse- 
power for 1 lb. of coal consumed per hour ; we hope for con- 
siderable improvement. Oil-engines give out one horse-power 
for less than 0*9 lb. of kerosene per hour ; we hope for very 
considerable improvement. 

45. Students ought to work many numerical exercises on me- 
chanical energy : w lb. of water raised vertically h feet, the energy 
is w/i foot-pounds. If time is given, find the work done per 
minute and divide by 33,000; this is useful horse-power. Divide 
this by the efficiency of a pump, and we have the actual horse- 
power which must be supplied to the pump. Or, if the w lbs. 
of water fall per minute through a turbine, the head available 
being h feet, we have w7i ^ 33,000 as the total horse-power, and 
the turbine will probably give out 70 per cent, of this usefully, 
0'70 or 70 per cent, being the efficiency of a good turbine. 

If there were no friction, a waggon of weight w requires to 



44 



APPLIED MECHA>*ICS. 



be palled with a force of w-^s up a road which rises 1 foot in 
everv s feet of its length. The frictional resistance to motion 
of a vehicle on a level road is usually stated as so many pounds 
per ton weight of the vehicla K there is also an incline we 
add the two tractive forces together — one for the incline without 
friction, the other to overcome fiiction on the level. The 
resistance in pounds per ton of a moving train (including 
engine) on the level is found roughly by adding 2 to one- 
quarter of the speed in miles per hour. Tiiis is for speeds 
greater than 20 miles per hour. At less speeds there is uuite 
a different law, which may for some trains and permanent 
wavs Ije indicated bv the f olio win sr fi^ires : — 



Speed in nules pear * 
honr. 1 

Eesistance in pounds | 
per ton. / 



20 


10 


2 

7 


5 
5 


10 
6 



A curved line adds i 2 f>er cent, to the resistance on the average 
English railways. The tractive force of heavy waggons on 
macadamised roads may be taken as 50 lbs. per ton, on paved 
roads 30 lbs. per ton, and on gravel roads as 150 lbs. per ton. 
These rules are good enough for academic exercise work. 

If the pull on a tramcar is recorded as the ordinate of a 
diagram, of which the al»scissa represents distance along the 
track, the area of the diagram represents to some scale the 
work done. The average height of the diagram represents the 
average pull p. This average pull P, multiplied by the whole 
length of the track, is the whole work done. In most practical 
cases the average can only be obtained by making the diagram, 
finding its area, and dividing by its length, just as we do with 
an indicator diagram of an engine. But there are cases where 
we can calculate an average. 

Example. — A chain of length I and weight tr/, with a 
weight w at the end of it, is to be wound up by a capstan ; 
what work will be done ? Obviously the average pull is w-}- 
half the weight of the chain, or w-|- J u^, and the total distance 
is /, so that the work done is w? + J ?r^. As a rule, it is \vise 
to plot the varying pull as the ordinate of a curve on square 1 
paper, as, without the aid of the calculus, one is apt to state what 
is the average pull without due thought. Thtis, if the abov»- 



APPLIED MECHANICS. 45 

mentioned chain varies in heaviness per foot, the average pull 
due to it is not half its weight. 

Observe that we ought to call the above the space average 
of a force. The space average of a force, multiplied by the 
distance, is the work done. 

A little consideration will show a student that a time 
average is a very different thing. 

46. According to the results of some experiments by Prof. 
R. H. Smith on the cutting of metal in the lathe without 
water or oil, the force on the tool is not much affected by 
speed. 

For both thin and moderately thick shavings at all speeds, 
feeds, and depths of cut, we may roughly take it that forged 
steel takes twice as much power to cut it as does cast ii^on; 
wrought iron takes one to one and a half times as much as cast 
iron. 

For broad thin shavings, cast iron required more cutting 
force than wrought iron. 

The force is neither proportional to the breadth of the 
shaving nor the depth, but it is more nearly proportional to 
depth than breadth. 

It is interesting to note that before these experiments it 
was usual in books to follow Weisbach in saying that for iron 
p :=/bd where/ was 50,000 lbs. per square inch. Smith's 

P 

experiments show that this rule is not true, and that — varies 

bd 

from 92,000 to 239,000. Possibly when it is discovered that 
there are other things to be done in college engineering labora- 
tories than to break endless numbers of specimens of metal 
with a 100- or 200-ton testing-macliine, we may have further 
experimental results on which practical engineers may rely. 

Professor Smith tried various depths, d inches, and breadths, 
b inches, of shaving from cast iron, wrought iron, and forged 
steel, in every case measuring p, the pressure on the tool in 
pounds, at the cutting edge. In almost every case we find from 
his numbers that the energy e usefully spent per cubic inch 
of metal removed, diminished about 30 per cent, as b was 
increased from "03 to '05, being about its minimum value when 
b was -05 ; but probably it does not increase much for greater 
values of b. The minimum values of E for various depths of 
cut were as follow : — 



46 



APPLIED MECHANICS. 



•05 


■056 


•135 


•056 


•03 


•056 


•06 


•06 


•14 


•056 


•02 


•056 


•04 


•056 


•06 


•056 



288 
690 
215 
570 
810 
250 
480 
705 



8,700 
7,700 
10,700 
13,000 
8,700 
18,700 
18,000 
17,700 



Cast iron. 

Wrought 
iron. 

Steel. 



With cast iron if 6 = '05, and probably for greater values of 
6, E is much the same for cuts of the depths -05 and •I 35 inch, 
being about 9,000 foot-pounds per cubic inch of m^tal removed. 

With steel, if b = '05, and probably for greater values of b, 
E is much the same for cuts of the depths ^02, -04, and 
•06 inch, being about 18,000 foot-pounds per cubic inch of 
metal removed. 

With steel, if 6= •OS, and probably for greater values of b, 
E is 10,700 foot-pounds for a cut of •OS inch depth, 13,000 
foot-pounds for a cut of '06 inch depth, and is practically the 
same as cast iron for a cut of '14 inch depth. 

EXERCISES. 

1. Two tons of rock can fall to a depth of 320 feet; find the work 
which it may do. Ans., 1,433,600 foot-lbs. 

2. In lifting- an anchor of 1 J tons from a depth of 15 fathoms in six 
miJiutes, what is the useful man-power, if a man-power is defined as 
3,500 foot-lbs. per minute? A)is., 14^4. 

3. The pull on a tramcar is 200 lbs. at an angle of 25° with the track ; 
what is the component in the direction of the track ? What work is done 
in a distance of 10 feet along the track? If the sj^eed is 4 feet per 
second, what is the usefully exi^ended horse -power ? 

Ans., 181-26 lbs. ; 1812-6 foot-lbs ; 1-318. 

4. What horse-power is inA'olved in lowering by 2 feet the level of the 
surface of a lake 2 square miles in area in 300 hours, the water being 
lifted to an average height of 5 feet ? Ans., 58-5. 

5. Taking the average power of a man as y\th of a horse-power, and 
the efficiency of the pump used as 0-4, in what time will ten men empty a 
tank of 50 feet x 30 feet x 6 feet filled with water, the hft being an 
average height of 30 feet? Ans., 21 hours 14 minutes. 

6. The diameter of a steam-engine cylinder is 9 inches, the length of 
crank 9 inches, the number of revolutions per minute 110, and mean 
effective pressure of the steam 35 lbs. per square inch; find the indicated 
horse-power. Ans., 22-3. 



APPLIED MECHANICS. 47 

7. One gas-engine uses 24 cubic feet of coal-gas, and another 98 cubic 
feet of Dowson gas per bour per useful horse-power ; what are their 
efficiencies ? The calorific powers of coal-gas and of Dowson gas per cxibic 
foot are 520,000 and 123,000 foot-pounds respectively. Ans., 0-159; 0-164. 

8. What would be the indicated horse-power of an Otto gas engine 
which has a piston 12 inches in diameter and a crank 8 inches long? 
The engine works at 150 revolutions a minute, and there is an explosion 
every 2 revolutions, the mean effective pressure in the cylinder during a 
cycle being 62 lbs. per square inch. Ans., 21. 

9. The average breadth of an indicator diagram for one side of the 
piston is 1-58 inches, and for the other side it is 1-42 inches, and 1 inch 
represents 32 lbs. per square inch. Piston, 12 inches diameter; crank, 1 
foot ; 110 revolutions per minute. What is the indicated horse-power ? 

Ans., 72-38. 

10. What must be the effective horse-power of a locomotive which 
moves at the steady speed of 35 miles an hour on level rails, the weight 
of engine and train being 120 tons, and the resistances 16 lbs. per ton? 
What additional horse-power would be necessary if the rails were laid 
along a gxadient of 1 in 112 ? Ans., 179-2 ; 224. 

11. In 10 find in each case how far the train woiild move after steam 
was shut off, assuming the above constant resistance and neglecting rota- 
tory motions. Find also the speed of the train after the latter had moved 
over a distance of 1,000 feet from the point where steam was shut off. 

Ans., 5,728 feet; 2,545*8 feet ; 31-8 miles per hour; 27-3 miles per hour. 

12. A flywheel weighs 2J tons, and its mean rim has a velocity of 40 
feet per second ; what is its kinetic energy ? If the velocity be redu(;ed 
3 per cent., what is the reduction in the kinetic energy ? If the kinetic 
energy be reduced by 10,000 foot-lbs., by how much is the velocity 
reduced ? In estimating the latter, why would it be wrong to subtract 
from 40 feet per second the velocity which corresponds to 10,000 foot-lbs. 
of energy in this flywheel ? 

Ans. 139,130 foot-lbs. : 130,900 foot-lbs. ; 1-5 ft. per second. 

13. A machine discharges n projectiles per minute, each of w lbs. 
moving with the velocity of v feet per second ; what is the actual horse- 
power ? If the efficiency of the machine is e, and it is di-iven by a steam- 
engine which uses w lbs. of steam per hour per horse-power given out by 
it, what is the total steam per hoiu^? If the engine is governed by 
thi'ottling, and if the total steam per hour follows the rule:— steam per 
hour z=za -{- b X brake horse-power (where a and b are known to us) , and 
if for half an hour n-^ projectiles are discharged per minute, and if for the 
next half hour «2 projectiles are discharged per minute, how much steam 
is used during the hour ? . wwv^ . wnssiP" . ^(?^J-^«2)wv^ 

'' ^x 66000 ' 66000^ ' ^ "*" 132000*7 * 

14. Town refuse is about /| a ton per unit -of the population per year. 
In the careful burning of 1 lb. of refuse, 0-5 lb. of water at 212*^ F. may 
be converted into steam at 212° F. If 1 lb. of coal is able to evajDorate in 
a good boiler 10 lbs. of water, how many tons of coal per year would pro- 
duce the same amount of steam as the refuse from 5,000,000 inhabitants ? 

Am., 5,000,000 x J x 0-4-flO, or 100,000 tons of coal per year. 
If we get one actual horse-power for 40 lbs. of refuse per hour ; if the 
engines drive pumps of 90,per cent, efficiency, pumping water to a reservoir ; 
if the water drives motors in the city with a total efficiency of 30 percent., 



is APPLIED MKCHANICS. 

working on an average, 2 hours per working day; what is the average 
horse-power supplied to each house if there are 500,0i)0 houses in the citv ? 
Ans., 5,000,000 xUx2240-=-40 or 42 x lO^" is the actual energy "in 
horse-power hours. The pumped energy is 3-78 x 10=* horse-power 
hours. The supplied energy is 1 134 x 10^ horse-power hoiu-s. The 
average total horse-power (2 hours a day for 313 days) is 1-8 x 10^, 
or 180,000, and per house it is 0-36 horse-power. 

15. The section of a stream is 12 square feet, the average velocity of 
the water is 2 feet per second ; there is an available fall of 25 feet ; -what 
is the horse-power available ? A turbine drives a dynamo machine which 
sends electric power to a motor at a distance. The efficiency of the 
turbine is 70 per cent. ; of the dynamo, 87 per cent.; ten per cent, of the 
energy from the dynamo is wasted in transmission, and the efficiency of 
the motor is 72 per cent. ; how much power is given out by the motor ? 
The voltage at the dynamo is 102 ; what is the current in amperes ? 

Ans., 68; 26-8; 303. 

16. Electric lamps giving 1 candle-power for 4 watts; how many 10- 
or how many 16 -candle lamps maybe worked per electric horse-power? 
The combined efficiency of engine, dynamo, and gearing being 70 per 
cent., what is the cajidle-power available for every indicated horse- 
power? ^/^5., 18; 11; 130-55. 

17. On a switchback the carriage is 966 lbs., neglecting friction ; find 
its kinetic energies when it is 5, 10, and 15 feet below its starting-point. 
And if the starting-point is 20 feet above datum level, ^\Tite out in two 
colimms its two kinds of energy at each point. If the above points are 
Q-l, 0-4, and 0-" of the distance along the track, and if loss of energy by 
friction is proportional merely to distance along the track, and if the 
carriage has to be lifted 1-6 feet at the end of each journey, find the 
correction in the kinetic energy at each place. 

Ans., Potential energies,' 14,490, 9,660, 4,830 foot-lbs. ; kinetic 
energies, 4,830, 9,660, 14,490 foot-lbs. ; corrected kinetic energies, 4,675-4, 
9,041-6, 13,407 foot-lbs. 

18. The calorific powers of 1 lb. of each of the following fuels are 
given in centisrade-pound heat units ; convert into foot-lbs. if 1 heat imit 
= 1,400 foot-lbs. :— charcoal, 7,000; coke, 7,000; coal, 8,800 to 7,330; 
wood, 4,200 ; and kerosene, 12,200. 

19. The pull on a tramcar was registered when the cax was at the 
following distances along the track: — 0, 200 lbs.; 10 feet, 150 lbs.; 
25 feet, 160 lbs.; 32 feet, 156 lbs.; 41 feet, 163 lbs.; 5Q feet, 170 lbs.; 
60 feet, 165 lbs. ; 73 feet, 160 lbs. ; what is the average (space) pull on 
the car, and what is the effective work done in pulling the car through 
the distance of 73 feet? Ans., 161 lbs., 11,800 foot-lbs. about. 

20. A chain hanging vertically 520 feet long, weighing 20 lbs. per 
foot, is wound up ; what work is done ? Ans., 2,704.000 foot-lbs. 

21. Four cwt. of material are drawn from a depth of 80 fathoms by a 
rope weighing 1-15 lbs. j)er linear foot ; how much work is done altogether, 
and how much per cent, is done, in lifting the rope ? What horse-power 
would be required to raise the material in four and a half minutes ? 

^«5., 347,520 foot-lbs ; 38; 2-34. 

22. (a) A cut of -06 inch depth is being made on a 4-inch wrought- 
iron shaft revolving ct 10 revolutions per minute ; the traverse feed is 0*03 
inch per revolution; the pressure on the tool is found to be 435 lbs.: 



APPLIED MECHANICS. 49 

what is tlie horse-power expended at the tool ? How much metal is 
removed per hour per horse-power ? 

Am., -1381 ; 98-28 cubic inches. 

{b) When the traverse feed is -06 inch per revolution, the pressure on 
the tool is found to he 570 lbs. ; find the horse-power, and the metal 
remoA'ed per hour per horse-power. Ans., •181 ; 150 cubic inches. 

(c) If the above horse-power is called the useful power u, and it is 
found that the actual horse-power given to the lathe is O'l-fTo u, find 
actual horse-power and metal removed per hour per actual horse-power. 
Ans., -3071, 44-2 cubic inches; -3715, 73'1 cubic inches. 

23. What is the kinetic energy of a tramcar mo-^dng at 6 miles per 
hour, laden with 36 passengers, each of the average weight of 11 stones ? 
Weight of car, 2| tons. What is its momentum? If stopped in 2 sees., 
what is the average force ? If the force is constant, this must also be the 
space average force ; find the distance of stopping if the force is constant. 

Ans., 13,400 foot-lbs; 3045-565 ; 1522-8 lbs. ; 8-8 feet. 

24. A ball weighing 5 ounces, and moving at 1,000 feet per second, 
pierces a shield, and moves on with a velocity of 400 feet per second ; 
what energy is lost in piercing the shield ? Ans., 4,076 foot-lbs. 

25. A fire-engine pump is provided with a nozzle, the section area of 
which is 1 square inch, and the water is projected through the nozzle 
with an average normal velocity of 130 feet per second; find (1) the 
number of cubic feet discharged per second, (2) the weight of water 
discharged per minute, (3) the kinetic energy of each pound of water as 
it leaves the nozzle, (4) the horse-power of the engine required to di-ive 
the pump, assuming the efficiency to be 70 per cent. 

Ans., (1) -9 cubic feet ; (2) 1-51 tons; (3) 262-3 foot-lbs. ; (4) 38-3. 

47. Bicycle problems. — When a man says that his bicycle is 
geared to D inches, he means that he advances tt d inches for 
one turn of his pedals. Let the diameter of the pedal circle 
be d inches. Let Wq be weight in lbs. of rider, Wq of machine, 
Wq-4-i^q=w. Let w be the uniform vertical force which the 
rider applies to each pedal alternately ; ii w is negative, it 
means that he is back-pedalling. Let F be the force in lbs. 
which would pull the bicycle along at the velocity of v miles 
per hour ; 2 lo d is the work done in inch-pounds by the rider 
in one revolution, and this is equal to tt D f. 

^ ' ^ , or pv -^ 375, the horse-power usefully expended.... (2) 

Uxvx 5,280 336 z; ,, . „ . ,. . . . 

^ = = n, the number of revolutions of a pedal per 

ttd X 60 D ' f f 

minute.... (3). 

[It is useful to remember that a machine geared to 56 inches 
goes at 10 miles an hour when the pedals make one turn per 
second.] 



50 APPLIED MECSA^flCS. 

The vertical force is not by any means constant in practice, 

nor indeed ought it to be : it is and ongbt to be greatest vhen 
the crank is horizoDtal. With proper ankle action the force 
is always somewhat in the direction of the circular path of the 
pedal, but for exercise work there is no harm in assuming a 
uniform vertical force o: in the down-stroke and no force in the 
up-stroke. In practice it is dilficult to avoid pressing on the 
pedal LQ its up-stroke. 

If Ff, is the value of f on a level ix^ad, then on a rising 
slope of 1 in s we have 

r = r, 4- J .... (4). 

If it is a descending slope, — is negative. F,-, may usually be 

taken as prop>oiiioiial to w. 

The following data were roughly measured by mysell 
They me good enough for academic problems. They suit my 
own bicycle on a good road. Students ought to obtain data of 
their own by careful meastirement. 

A rider weighing 127 lbs. on a cycle weighing .33 lbs. (or 
w=160 lbs.) finds that on a descent of 1 in 80, with his feet off 
the pe<:3als, he is ju^t able to get on very slowly but steadily ; 
on a long slope of 1 in 40 his steady speed (feet off p>edals) was 
9 J miles i>er hour ; on a long slope of 1 in 20 his steady speed 
(feet off pedals; was 20 miles per hour. In these three cases 
the values of Fy were : — 

W r. W ^ W o 

— or 2, — or 4, ^ or 8. 

SO '40 20 

We need careful expeiiments ; and it is not of much use 
speculatiivg on the probable law of resistance of a safety bicycle 
with pneumatic tyres on a certain kind of road. A constant 
term for quasi solid friction, and a term (the most important at 
high speeds) proportional to the square of the speed relatively 
to that of the atmosphere for air resistance : these we ought to 
have. Resistance due to unevenne&s of the ground would be 
constant if a certain kiud of unevenness were to repeat itself 
at intervals so far apart on the road that the vibration due to 
each had time to die away before the next ; but speculation is 
vague, especially as the kind of vibration will often depend on 
the velocity. In our present state of knowledge we cannot be 
far wrong in assuming 

p^ = w (o + Zt -h ct^). 



APPLIED MECHANICS. 



51 



If the above values of Fy are correct at the given speeds, 
y— 0, 9|, and 20 miles per hour, we find 

•^ 80 V 20 ^ 200 y' ^ ^ 

as a formula which represents our experimental data well 
e'iough for exercise purposes. 

Hence, going up a slope of 1 in s we have 



= -( 

80 V 



t; j^ 8_0 
20 200 s 



tor this bicycle. When w= 160 lbs. 

F = 2+ 1+J^ + ^ .... (6). 
^ 10^ 100 5 ^ ' 

Example 1. — What horse-power is expended in going at 

12 miles an hour on a level road ? Here =0. 

s 

F = 2+^+— = 4-64 lbs. 
10 100 

The speed is 88 v or 1^056 feet per minute, and the horse power 

_ 1,056X4-64 ^ Q.-^^g 

33,000 

Example 2. — At 12 miles an hour, going up or down an 

incline of 1 in 60, what is the useful horse-power^ 

F ^ 4-64 -\-\ — 7-31 or 1-97 lb. 
— 60 

Horse-power up = 1,056 x 7-31 ^ ^.^34^ 
33,000 

Horse-power down = M^^JLL^^ = 0-033. 
^ 33,000 

EXEECISES. 

1. In Examples 1 and 2, what force, w, does the rider exert upon 
his pedal if his bicycle is geared to d = 60 inches and the diameter of the 

circle is 13 inches ? 
Ans.^ On the level, w z= 33-6 lbs. ; going- up, w = 53 lbs. ; gomg 
down, tv r= 14-3 lbs. 

2. On what slope downward would the velocity of 12 miles an hour be 
steadily maintained, feet off pedals ? Atis. , 1 in 34|. 

3. Gjoing down a slope of 1 in 30 at 8 miles per hour, what is the force 
on the pedal ? A>is., — 12-30 lbs. 

The minus sign means that the rider must back-pedal. 

4. If a rider whose weight is 157 lbs., back-pedals on this same bicycle 
with a force of only 10 lbs., down a slope of 1 in 25, what is his velocity f* 

Ans., 13 '6 miles per hour. 



52 APPLIED MECHANICS. 

The foUo-^-mg example is for students who can integrate : — 
The resistance to motion being f z= a -\- bv -\' cv'^ on a road 
■svhich does not alter in character for a whole joTimey, compare the 
work done in going over a certain distance, I — first* at a constant 
speed, Vq ; second, at the same average speed, but varying according 
to the law v = v^ -\- f sin. qt. In both cases the time is the same 
if the journey is just so long as to be finished by the rider in the 
same state as to speed, etc., with which he starts. 



Jp..=J.|rf.=Jp....=J( 



(av + bv- + cv^) dt, 

the work done. If we calculate this for a time, t, where q = 27rT, 
it is just as good as for any niimber of such periods. The value 
divided by t gives the average work per second, or 

av, + bv,^ + a-o3 + i i/ + I cv,P ; 

whereas, if / is and the speed is constant, we have the average 
"work per second av, + bv,- + cv^^. Hence the fractional increase of 
work is {bf -f ^cvj^) / 2 {av, + ^I'o" + '^^''o^)- Thus, taking our 
values of Art. 47 for a weight of 160 lbs., 

F = 2 + - V -\- V' A- 

10 '100 ^ s 
If the road has everywhere an upward slope, 1 in 5, we take 

^ = 2 + ^2; 

sl 

but we had better take a level road, so that a = 2. In anv case, 

b= % cz=z -01. 

Taking an average speed of 10 miles per hour and 

v = 10 4- 3 sin. qt, 

so that the speed fluctuates between 13 and 7 miles per hour, we 

have a fractional waste of po'^er 

3 i -f 90c 3 

— or - : 

20 a + 10* + 100c 80 

that is, the power at constant speed being 80, the average power at 

varying speed is 83. 

48. Much of what we have given may be said to be mere exer- 
cise work in the use of formulce. But we hope that it is much 
more than that, and that students are getting to understand 
how the formulEe are derived. Much of this book is devoted 
to the explanation of how these formulae are derived, and 
somewhat similar exercises will be given later. Our object 
has been to familiarise students with the notion of the 
quantitative transformation of energy. The subject of this 
book is almost altogether the study of energy and momentum. 
We close this introductory part of our subject with a few 
problems on the hydraulic transmission of power and the 
propulsion of ships. 



APPLIED MECHANICS. 



53 



Foi- academic exercises it is sufficiently correct to say (see Art. 69) 
that the energy wasted per pound of water flowing in a pipe is 
experimentally found to he h times its kinetic energy, where A: has the 
following values: — InZfeet of pipe of diameter ^feet, X; = -0232 I'jd ; 
entrance or exit hy cylindi'ic pipe to or from a reservoir, ^ r= 0-5 ; 

bend in a pipe, ^= < -131 + 1-847 / - W^ | a, where a is the 

fraction of two right angles through which the bend extends, 
D the diameter of the circle of which the centre line of the pipe 
is part and d the pipe's diameter. Probably only the first of 
these values of h is fairly correct. 

Exercise.- — Prove that if the horse-power, h, enters a straight 
pipe as pressure water, the waste power, w, is '00374 l^jp^d^, 
where p is the pressure at the entering end in pounds per square 
inch. Notice that the fractional loss of power is the fractional 
loss of pressure. 

If v is the voltage and a the current in amperes, the horse- 
power delivered electrically to a condiictor is v a watts (746 watts 
are equal to 1 horse-power). The loss in the conductor is a^ k 
watts, if R is its resistance in ohms. Hence, if h is the horse- 
power sent in, the wasted power is w =: 746 h^ r/v^. 

As E, = - X '044 if the copper conductor is n miles long (going 
a 
and coming, so that the distance is \n miles) and a square inches 

V? n 
in cross-section, w =r 32-7 — ., -• Notice that the fractionalloss 
v^ a 

of power is the fractional loss of voltage. 

Exercise. — Prove that at an entering pressure of 700 lbs. per 

square inch, if we admit the following amounts of hydraulic 

power, we have the following amounts of waste power. Also find 

the values in the table for a conductor. 



H 


At pressure of 700 lbs. per 

square inch. 

Horse-power lost in one mile. 


At 700 volts. 
Horse-power lost in one mile. 






Conductor 
0-25 square inch 
in cross-section. 




The Power 
sent in. 


6-inch pipe. 


3-inch pipe. 


Conductor 
0-125 square inch 
in cross-section. 


20 

50 

100 

200 

300 


0-23 
1-84 
14-72 
117-8 


0-9 

7-4 
59 


0-67 

2-67 

10-68 

24-03 


0-34 

1-34 

5-34 

21-36 

48-06 



Up to the highest usual speeds of commercial ships we 
may assume without great error that, for vessels not dissimilar 



54 APPLIED MECHANICS. 

in form and cliaracter and going at the nsnal speeds, the 
indicated horse-power is h = d| t^ -^ <•, where d is the dis 
placement in tons and v is the speed in knots and c is a 
constant, which for many classes of vessel may be taken as 
not Tery different from 240. 

Exercise 1. — ^What is the indicated horse-power of a vessel of 1,330 
tons moving at a speed of 12 knots, if it obevs the above mle * 

Ans., 871. 
Eiireiie 2. — ^If a vessel of 1.720 tons moves at 10 knots whai 
its indicated horse-power is 655, what is the Talne of e in such a da^ of 
res^l? ^iJM., 219. 

The resistance to the motion of a ship is oondd^ied to be made 
np of two parts. 1. The skin friction in ponnds, s:=/a v *, where 
V is the speed in knots, » is 1 '83 for Tarnished or painted wooden 
modds or dean iron ehips, a is the wetted soea in square fe^, 
/ is *009 for diipe of over 200 £eet long, and -012, -0106, -0096 
for ship lengths of 8, 20, and 50 fedt. At speeds of 6 to 8 knots in 
ordinary ve^els this skin resistance is ahont 80 or 90 per cait. of 
the whole ; at hi^ speeds it is abont half the whole. 

2. A residuary resistanee doe to the fact that eddies rhe 
anaHer part) and waves are prodneed. Eddy resistance :5 :':i -> : 
not to be more thffla 8 per cer.^ :f tie ;lir- rr : ' - ^ t _ 

speeds. It is mainly cansei : 7 ": 1 — r- - : :.1t :: 
]£ two perfectly shmhtt- sL: - ::.! : . 1 

i^ at speeds t and f \/I/^ ' : 

ipeeds, the residuary resistar : _ 

The skin r^istanc^ Sj ;i^ : - -i^ . ' - 

calcnlated from Fronde's 1^ : . r. 11- : ^^ 

ther^istance in ponnds of 1 : ;: 1 n^ :: - i 1 ir^ 

square feet, v its speed in 1 i^ : ji : : e 

and lengih of a model wli r_ 

draught when the model i^ _: :. 

r knots, where v:r:: v^l; \ '.. l "^ — ^i i;: iv^-. »; -.i_ \_- 
above that 

^ r— -009 Av^"^ I lib 



= r^r— -009 A v^^ nib 



a the ship is more than 200 fe^ k>ng, and the model is from S to 
30 feet long. 

Example. — ^Before building a vessel 400 feet long, of wdted 
surfac-e 26,000 square feet, we wish to know k, its resistance, at 
V = 12 knoisL A model is made 10 feet hmg, it is drawn at a 
speed of 12v/40 or 1-9 knots in the tank, and its r^istanee r 
is found to be 0-9 lb. We find r to be 39,720 lbs. 

Prove that k in pounds x v in knots -r- 325 = utilised horse- 
power. In this case we find 1,463 horse-pow^^. The indicated 
power win prolably be more than 3,000. 

The vagueness of our knowledge as to ihB probable loss of 
power by friction, makes any attempt to calculate s. for the above 



APPLIED MECHANICS. 55 

purpose rather useless, and the better use of the tank would there- 
fore seem to lie in helping- to improve a particular class of vessel. 

The following- great simplification has recently been tried by 
Colonel English. Suppose an existing vessel to be run at various 
speeds and its indicated horse-power noted. Xow, assume that the 
effective horse-j)ower in a new ship will be the same fraction of the 
indicated that we take it to be in the existing- ship — say one-half. 
Find the resistance of the existing- ship at the sj)eed v^. We wish 
to know the resistance of the new ship at the speed Vg. We only 
need to. compare the wave and eddy resistances, which we shall 
call Wj and w.^. Make two models, one of the existing ship and 
one of the ship being- designed. Let the values of v, d, s 
L, w for the two ships and the two models be indicated by 
capital and small letters, the existing ship and its model having 
the affixes ^. s is skin friction ; d is displacement, which in similar 
shi^DS is proportional to the cubes of the lengths. 



Let.. = v.(|)V, ., = .(1^)\ 



and let 



that is, make the models of such sizes that V2 and V2 as well as v^ 
and Vj_, are " corresponding speeds," and yet that the speeds of the 

two models shall be the same. In fact, — = —(-]• Now let 

the two models be towed from the two arms of a lever whose 
fulcrum may be adjusted and the ratio of the resistances, n, of 
the second to the first may be measiu-ed. Note that we need only 
find this ratio — a much easier thing to do than to find either 
resistance. Show that the total resistance of the new ship is 



(?y\ 



i + 7- ("^1 " «2) 



66 



CHAPTER IV. 

FRICTION. 

49. We have said that the mechanical principles which must 
be studied by the young engineer are few in number, but they 
must be very familiar to him. It is not well to say that sltij 
one method of study is more important than another, the fact 
being that a student must not only study in the workshops 
and drawing-office, but he must read, work numerical exercises, 
and make a great many quantitative laboratory experin;ents 
to illustrate these principles. Our aim is to get students to think, 
and it is astonishing how difficult it is to effect this object. 
We cannot easily get students to wrangle over these subjects. 
We have few pretty lecture experiments. Even in chemistry 
and experimental physics, pretty lecture experiments are not 
very effective in causing students to think. Students will 
think about things that they do. Hence it is that boys should 
be allowed to chip and file metals, and to pare and cut wood. 
Merely in learning how to hold a chipping-chisel or in setting 
a plane iron, a student must think about the properties of 
materials and forces. Country boys who make their own things 
have a great advantage over town boys who buy their things 
in shops. In the mechanical laboratory, I find that even the 
dullest student begins to think for himself if he is not too 
much spoon-fed ; and if his difficulties are not cleared away by 
some wretched routine system of laboratory work being adopted 
by cheap laboratory instructors, the fundamental principles of 
mechanics will become part of his mental machinery. 

It is not necessary to illustrate everything ; a few things 
carefully done in the laboratory are better than many. For 
example, let the triangle of forces be illustrated in its simplest 
form. The principle is : — If three forces act on a small body 
and just keep it at rest, then if we draw on a sheet of paper 
three straight lines parallel to the directions of the three forces, 
and let them form a triangle in such a way that arrow-heads 
representing the directions of the forces go round the triangle 
circuitally, it will be found that the lengths of the sides 
of the triangle are proportional to the amounts of the forces. 
Fig. 19 shows how the strings, pulleys, and the smooth ring p 



APPLIED MECHANICS. 



57 



are used. I am in the habit of 
using three scale - pans with 
weights in them, and I measure 
beforehand the Aveights of the 
scale-pans themselves. Put al- 
most any weights in at random 
(only you will tind that any two 
must be greater than the third), 
and let P find its position of 
equilibrium, and you w411 find 
the rule to be nearly true every 
time. Also for the same set of 
weights you will find that there 
is a small region within which, anywhere, the 
centre of the ring p may be placed without dis- 
turbing the state of equilibrium ; this is owing 
to the friction of the pulleys. The polygon of Fig. 19. 

forces is also easily illustrated. (See Fig. 20.) 

50. Our one Theory insufficient. — No man can make these 
trials without finding that there is a great deal to be observed 
beyond what his 




teacher 



his 



book has taught 
him. A force 
has been repre- 
sented by the 
pull in a string 
passing over a 
little pulley with 
a weight at its 
end. He finds 
that as his pul- 
ley works more 
easily, and as its 
pivots are better 
oiled, his illus- 
tration of the 
law is better and 
better ; in fact, 
he finds that the 
pullinastringis 
not exactly the 




Fig. 20. 



58 



APPLIED MECHANICS. 





Fig. 21. 



same on the two sides of a pnlley. If he takes 
one pulley and one string and two weights, 
called A and b, Fig. 21, at its ends, he will 
find that there is equilibrium even when 
the two weights are not exactly equal. If 
A is slightly greater than B, and he further 
increases the weight of a till it is just able 
to overcome b, then the difference between 
the weights represents, in some fashion, 
what may be called the friction of the 
pulley. If now he increases both the weights 
which he uses, he will find that the fric- 
tion is propordonately increased, and he 
will get to understand why we so generally 
find in machinery that there is a law, "fric- 
tion is proportional to load.' This law 
is not quite true, but it is sufliciently true 
to be of great value to engineers. Again, 
he sees that this friction, which is a 

resistance experienced in the rubbing together of any two 

surfaces, is a force which always opposes motion, always acts 

against the stronger influence. Suppose, 

for example, that he found that a 

weight of 5*1 ounces was just able to 

overcome a weight of 5 ounces ; he 

will find that a weight of about 4*9 

ounces will just be overcome by a 

weight of 5 ounces, and that there is 

equilibrium with 5 ounces and any 

weight varying from 5*1 to 4 '9. Fric- 
tion always helps the weaker forces to 

produce a balance. 

51. Law of Work. — Take any 

machine, from a simple pulley to the 

most complicated mechanism. Let a 

weight, A, hung from a cord round a 

grooved pulley or axle in one part of 

the mechanism, balance another weight, 

b, hung from a cord round another axle 

or pulley somewhere else. In Fig. 22 

we have imagined that the mechanism 

is enclosed in a box, and only the two ^^g- ^- 




A.PPL1ED MECHANICS. 59 

axles in question make their appearance. Now move the 
mechanism so that A falls and b rises steadily. Suppose 
that when A falls 1 foot b rises 20 feet, then if there were no 
friction in the macliine a weight at A is exactly balanced by 
one-twentieth of this weight at b. This is the law which you 
will find proved in books on mechanics. The reason why it 
is true is this. The work or mechanical energy given out by a 
body in falling is measured by the weight of the body multiplied 
into the distance through which it falls. It is in this way that 
we get the energy derivable from the fall of a certain quantity 
of water down a waterfall, and it is in this way that we find 
out Avhether a certain waterfall gives out enough power to 
drive a mill. Similarly the energy given to a body when we 
raise it, is measured by the weight of th-e body multiplied by the 
vertical height through which it is raised. Now, every experi- 
ment we can make shows that energy is indestructible, and 
consequently, if I give energy to a machine, and find that none 
remains in it, that there are no means there of converting 
mechanical energy into heat by friction or into any other form 
such as electiical energy, or vice versd, and no storage or 
unstorage of energy, as by lifted weights or the coiling of 
springs, or by increasing or diminishing the kinetic energy of 
anything (this is why I pre-suppose uniform velocity in A and 
b), then all the energy given to the machine must be given out 
by it. This is often what people really mean when they say 
that their machine is supposed to have no friction. 

Therefore the energy given out by a in falling must be 
equal to the energy received by B in rising; and as A falls 1 foot 
when B rises 20 feet, the weight of A must be twenty times the 
weight of B. If, then, there were no friction in the machine, 
and if a weight of 20 lbs. were hung at A and a weight of 1 lb. 
at B, we should find tliat if we start a downwards or upwards 
there will be a steady motion produced. Any excess at a will 
cause it to overcome b, the weights moving more and more 
quickly as the motion continues. 

Now, in our machine. Fig. 22, we can always find by trial 
what is the velocity ratio — that is, the speed of b as compared 
with the speed of a — and this is usually called the mechanical 
advantag'e when there is no friction, I have chosen a machine 
in which I suppose that if A has a uniform motion, so has b. 
But if when a is uniform, b is not uniform in its motion, then 
the velocity ratio for any particular position must be measured 



60 APPLIED >]ECHA>'ICS. 

during exceediiiglj small motions, as, after a little motion every- 
thing alters. Let us continue to suppose that the velocity 
ratio does not alter. Kow, when we try to balance a weight 
at B by a weight at A, we find that the above relation is quite 
untrue. Hang a weight of 1 lb, at b, hang a weight of 20 lbs. 
at A, there is certainly a balance ; but when we have somewhat 
less or more than 20 lbs. at A there still is balance. The 
reason for this is that there is friction in the mechanism, and 
this friction always tends to resist motion, always acts against 
the stronger influence. 

52. Effect of Friction. — We proceed to find out in what way 
friction modifies the law given in books. You must make 
actual e.xperiments with some machine if you are to get any 
good from your reading. Hang on a weight, B, and find the 
weight, A, winch will just cause a slow, steady motion. Do 
this every t:me when a number of different weights are 
placed at b. Kow suppose you have measured the velocity 
ratio — that is, suppose you find that B rises four times as 
rapidly as a falls. Then, according to the books, there would 
be an exact balance if a were four times the weight cf b. 
On actual trial, however, I find in a special case the following 
table of values : — 

A overcomes B wlieii 
A is 23 "4 ounces and B is 5 ounces 



44-7 




10 


65-4 




15 


86-8 




20 


107-5 




7 25 


128-8 




30 


149-6 




, 35 


171-0 




, 40 



But if there had been no fidction in the first experiment, A 
would have been 20 ounces instead of 23-4, hence the friction 
is represented by this 3-4 ounces. For every experiment let 
this be done, subtract four times B from A and call this 
difference the friction, Kow how shall we compare this 
friction with the corresponding load ? 

53. The Use of Squared Paper. — And here we come to a 
matter of the greatest importance to the practical man. How 
do we practically compare two things whose values depend on 
one another ? How do we find out the law of their de- 
pendence ? It is a strange fact that there should be a class in 



APPLIED MECHANICS. 



61 



^ H- 


^ ^ ' 


\ 


>^' 


\ 


V 41 


A 


^ it 


N 


V 


\ 


\ 


Jx\ 


V 4: 


^ 


\ 


\.- 


^ 


\ 


\ 


«v\ 


V 


\ 


\ 


\vcr 


\ 


5 


\ 


3 


c^V 


\ 


\ \ 


it \ 


aV 


A 


\ 


Jl it 3 


^ 



Fig. 23. 



the community who have a little difficulty in manipulating 
decimals in arithmetic, but it is almost a stranger evidence of 
neglected education that so many people should be ignorant 
of the great uses to which a sheet of squared paper may be put. 
A sheet of squared paper can be bought very cheaply. It haa 
a great number of horizontal lines at equal distances apart, and 



6^ 



APPLIED MECHAXICS. 



these are crossed by a great number of vertical lines of the same 
kind, so that the sheet is covered with little squares. This 
sheet will enable me first of all to correct for errors of observa- 
tion in the above series of experiments ; and, secondly, to 
discover the law which I am in search of. A miniature 
drawing is shown in Fig. 23, many lines being left out because 
of the difficulties of wood-catting. At the bottom left-hand 
corner I place the figure 0, and I write 10, 20, etc., to indicate 
the number of squares along the line, a. Instead of 10, 20, 
etc., I might write 1, 2, etc., or 100, 200, etc., according to the 
scale I am going to use. Indeed, on account of the friction 
being so much less than the weight a with which it is to be 
compared, I number the squares along the vertical line F by 
1, 2, etc., instead of 10, 20, etc. We can employ any scale 
we please in representing any of the things to be compared, 
and it is usual to multiply all the numbers of one kind by some 
number, so as to represent all our experiments on one sheet of 
paper, and on as much of this sheet as possible. Having sub- 
tracted four times b from a, I find the following numbers : — 



A. 

23-4 
44-7 
65-4 
86-8 



Friction. 


A. 




Friction. 


3-4 


107-5 . 


, , 


. 7-5 


47 


128-8 . 


. 


. 8-8 


5-4 


149-6 . 


, , 


. 9-6 


6-8 


171-0 . 


. 


. 11-0 



I now find on my sheet of paper the point p, which is 23-4 hori- 
zontally and 34 vertically, and mark it with a cross in pencil. 
Q is 44*7 horizontally and 4 "7 vertic-ally, and so for the others. 
The last point, w, is 171 horizontally and 11*0 vertically. We 
guess at the decimal part of a small square. The point p 
re})resents the two numbers of my first experiment, and every 
other point represents the two observations made in one 
experiment. Kow we are certain that if there is any simple 
law connecting load and friction, the points p, Q, to w, lie in a 
simple curve or in a straight line. You see that in this case no 
curve is needed to suit the points ; Ave assume that they would 
lie in a straight line, only that we made some errors of observa- 
tion. You must now find what straight line lies most evenly 
among all the points ; this you can do by means of a fine 
stretched string, and the line m n seems to me to answer best. 
It tells me, for instance, that when A is 44*7 the friction is 
really 4*5, instead of 4-7. Take any point in the line, its 



APPLIED MECHANICS. 63 

vertical measurement gives me the true friction corresponding 
to a load represented by its horizontal measurement. Thus, 
for instance, you see that friction 5*0 corresponds to 
load 55. 

This is a simple way of correcting errors made in experi- 
ments, but you cannot hope to understand much about it till 
you actually make experiments, and use the squared paper. 
You will find the matter all very simple when you try for 
yourself j my description of it is as complicated as if I were 
teaching you by mere words to walk, or to bicycle. 

If at any time you make a number of measurements of 
two variable things which have some relation to one another, 
plot them on a sheet of squared paper, and correct by using a 
flexible strip of wood or a ruler, to draw an easy curve or a 
straight line so that it passes nearly through all the points. 
If the line is straight, the law connecting the two things will 
prove to be a very simple one. In the present case it means 
that any increase in the load is accompanied by a proportionate 
increase in the amount of friction. Thus, when the load is 0, 
the friction is 2-3; when the load is 100, the friction is 7-2. 
That is, when the load increases by 100, the friction increases 
by 4-9, so that the increased friction is always the fraction, 
•049, of the increased load. In fact, it is evident that we can 
calculate the friction at any time from the rule 

Friction = 2-3 + '049 A. 

That is, multiply the load A in ounces by '049, and add 2-3, 
the answer is the friction. 

54. Law of Friction. — Our result is that the total friction is 
equal to the friction 2*3 of the machine unloaded, together 
with a constant fraction, -049, of the load. ISTow when a 
similar series of experiments is tried on any machine, be it a 
watch or clock, or be it a great steam-engine, we always find 
this sort of simple law. 

If you clean all the bearings or pivots, or if you use a 
different kind of lubricator, you will get other values for the 
two numbers in the above rule, but the law will remain of the 
same simple kind. I find it nearly impossible to get my pupils 
to believe that rough and ru^ty old machines, such as screw- 
jacks or hydraulic jacks, which have been long in use, are 
far more instructive to study than beautiful, specially made, 
frictionless machines. In my laboratory, now, at Finsbury, 



64 APPLIED MECHANICS. 

there is an ideal screw-jack ; the weight p (Fig. 40) is not a 
single weight, but two equal ones at the ends of two cords 
which pass round two equal grooves and produce a true COUple 
in turning the screw. I prefer the rough old thing previously 
in use, just as I prefer the cheap Attwood'S machine (Fig. 163), 
which used to be employed in my laboratory, to either of the 
two elaborate machines which are now in use. The pulley of 
Fig. 21, which I use, is out of balance and badly made; it 
gives ever so much more instruction than if it were so 
expensively and correctly made that the friction which one 
wants to measure had almost disappeared. My laboratory 
crane is a model crane, much too carefully constructed ; but 
my hydraulic jack and differential pulley block are the real 
things, made ''for human nature's daily use." Our object is 
not to find out how to make a machine with the most friction- 
less bearings ; else wo should find it instructive enough to 
work experimentally with ball bearings, such as are used in 
cycles (Art. 70), and with friction wheel bearings. Again, 
a student is told to judge with his eye as to whether a weight 
A is falling steadily, with a uniform velocity. Let him find 
out for himself how much of the steadiness is due to irregu- 
larities in the rubbing surfaces of the machine, and how, much 
can be altered by altering the weight. Do not spoon-feed him. 
Let it be a discovery of his own that his eye is somewhat 
defective as a speed-measurer ; he will be led to suggest plans 
for more accurate working if you refrain from forcing upon 
his attention your elaborate plans. Lideed, your electric and 
other contrivances for measuring velocity may be so elaborate 
as to hide altogether from a student the main object of his 
experiments ; just as when a young student works with a 200- 
ton testing-machine, it is almost impossible for him to think 
of the little specimen of material which is being tested, the 
testing-machine itself takes up so much space. 

55. Force of Friction. — I have in all this used the term "fric- 
tion," or the term " effect of friction," to mean the difference 
between the weight which would balance another through the 
mechanism if there were no resistance to the rul)bing of 
surfaces, and the weight which will just overcome the other 
when there is such resistance. Observe that there is a great 
difference between the cases, A overcoming b (Fig. 21), and b 
overcoming a. What we have called friction is due to the 
rubbing at all sorts of surfaces in all sorts of directions, at all 



APPLIED MECHANICS. 



65 



n-iiiiiiiiililii!,^^^^^^^^^^^^^^^^ 




sorts of velocities under all sorts of pressures, and we are led 
to study it in its simplest form, where at one part of a pair of 
surfaces the rubbing is exactly of the same kind and in the 
same direction as at another part ; so that we may speak of the 
resultant force which resists motion as the force of friction. 
Experiments may be made upon the apparatus shown in Fig. 24, 
where A B represents a table, the upper level surface of which 
is wood, iron, brass, or other material to be experimented 
upon. We usually experiment on smooth surfaces, c is a 
little slide made of any material whose coefficient of friction 
with the table we wish to find. Different weights may be 
placed on it. The weight of the slide, together with the 
weight lying upon it, is the total force (r) pressing the two 
surfaces together, c is pulled by the weight, w, hung from a 
string, passing over a pulley working on very frictionless 
pivots. The weight, w, which will just cause the slide to keep 
up a steady motion on the table, is taken as a measure of the 
friction. Of course, however, it really includes the resistance 
of the pulley, but this is usually neglected, as we know from 
previous experiment that it is small. It is found necessary to 
start the slide by giving a little jerk to the arrangement, as 
the friction when the slide is motionless is found to be some- 
what greater than when it is moving. This is one of the most 
instructive experiments which can be made in mechanics, and 
I hope that every reader will make a series of observations. 
Let him correct his results by means of 'squared paper, and he 
will find it nearly true that the friction is a constant fraction 
of the force pressing the surfaces together. This fraction is 
called the coefficient of friction and usually denoted by /x. I 



Q6 



APPLIED MECHANICS. 



give its value for a few surfaces, but a student had better 
depend upon the values which he himself arrives at ; they will 
not be the same ; thej may differ gi'eatly from these. 

Oak on oak, fibres parallel to direction of m-tion . . 0-48 

., „ perpendicular ,, ,, . . 0-34 

„ ,, endwise „ ,, . . 0-19 

Metals on oak ,, parallel ,, „ . Oo to 06 

Wrought iron on wrought uon^ wrought iion on cast 

ii-on . 018 

Cast iron on cast u'on . . . . . O'lo 




Fis. 25. 



A layer of oil or other lubricant between the surfaces will 
greatly reduce the friction. Figures given for the coefficient 
of friction when a lubricant is used are. however, very greatly 
misleading, but for student's exercise work p. may be taken any- 
thing between -04 and -01 for sperm oil, between not too gener- 
ously btit continuously lubricated surfaces, being twice as much 
for greases as for sperm oil. 

It is very interesting, after determining the coefficient in 
the case of a certain pair of materials, to diminish the size of 



APPLIED MECHANICS. 67 

the slide. You will find that unless you diminish it so much 
that the pressure actually alters the surfaces in contact from 
being quite plane you will get pretty much the same result. 
You will also find that, whether the motion of the slide is 
quick or slow, if the weight maintains the motion steady 
when it is slow it will also maintain it steady when quick: 

If, instead of using a cord and weight, we move the slide 
by tilting the table more and more from the horizontal (see 
Fig. 25), the slide getting an occasional shove to start it, let the 
inclination of the table be found in degrees when the weight 
of the body itself is just able to keep up a steady motion. The 
tangent of the angle of inclination of the table when this 
occurs can be found in a book of mathematical tables ; it proves 
to be equal to the coefficient of friction. This method of ex- 
perimenting is much easier and is more exact than the other, 
but it is not so instructive for a beginner, (See Art. 90.) 

56. Loss of Energy due to Friction. — In the simple case 
with which we began Art. 50, the difference of pull in a cord 
on the two sides of a jDulley was what we called the friction 
of the arrangement, whereas we see that the friction takes 
place at every point where rubbing occurs, not only at the 
pivot but even in the fibres of the cord itself, and the force 
at one point may be very different from the force at another 
pohit. Again, any force acting on the cord has a greater 
leverage about the axis than any of the forces of friction has. 
The real connection between the two things is, then, this : 
what we have generally called " the effect of friction," or "the 
friction of the arrangement,'^ multiplied by the velocity of the 
cord on which it is measured, is equal to the sum of all such 
products as the friction at any point in a rubbing surface, 
multiplied by the velocity of rubbing. In fact, if the Nveight a 
in falling causes the weight b to rise, the work done by a is 
greater than the work done on b by an amount which is called 
the work lost in friction, and this is the work done against the 
forces of friction at all the rubbing surfaces. 

If we know the force of friction at any place, in jjounds, 
and the distance, in feet, through which this force is overcome — 
that is, the distance through which rubbing has occurred — the 
product of force by distance measures the work or energy spent 
in overcoming friction, in foot-pounds. This energy is all 
wasted, or rather, it is all changed into heat and does not come 
out of the machine as mechanical work, the shape in which it 



68 



APPLIED MECHANICS. 




Pig. 26. 



was when we put it into the machine. And inasmuch as no 
machine can be constructed which will move without friction, 
we never get out of a machine as much mechanical work as we 
put into it. 

57. Friction at Bearings of Shafts. — At almost every 
rubbing surface which you can consider, tlie force of friction is 
different at every point of the surface, and it is generally acting 
in different directions at different points. Consider, for example, 
a horizontal shaft and its bearing (Fig. 26). The force of 
friction at c, per square inch of area of 
rubbing surface, is probably not the same 
as at A. A very little difference in the 
sizes of the journal and step will cause 
a considerable difference in the pressure per 
square inch at c or at a. Xow the 
force of friction at c, multiplied by the 
velocity of rubbing, gives the work or 
energy lost per second in friction at c ; 
and this, added to the energy lost at every other place 
where rubbing occurs, gives the total loss of energy per 
second at all the points. It is not, then, a simple matter to 
investigate the force of friction at every point of such a 
bearing ; and the rigidity of the metal, arid a number of other 
important matters, not to speak of the nature of the lubricant, 
must be taken into account in investigating the force of 
friction everywhere when the shaft is transmitting different 
amounts of power. As we have already seen, however, experi- 
ment shows that the energy lost in friction for a certain amount 
of motion increases proportionately with the energy actually 
transmitted by the shaft. Keeping in mind, then, the general 
law — " the force of friction is proportional to load "' — it is easy 
to see how to reduce the friction al loss in any machine. For 
instance, when a wheel is transmitting power, the load on the 
rubbing surfaces of its bearings or pivots depends on the power 
transmitted. Now, the actual force of friction at the rubbing 
surface is about the same, whatever be the size of the bearing ; 
but the distance through which rubbing occurs when the wheel 
makes one revolution is less as we have a less diameter of bearing ; 
in fact, the force of friction, multiplied by the circumference of 
a cylindric bearing, is the energy in foot-pounds lost in one 
revolution. Our rule is, then, to make this diameter as small 
as possible, consistently with sufficient strength. The wheel of 



APPLIED MECHANICS. 69 

a carriage is made large, and the axle, where rubbing occurs, 
is made as small as possible, because in this way the carriage 
moves over a great distance for a small amount of rubbing. 
There is another reason, however, for the use of large wheels 
in carriages on common roads — namely, their requiriDg a less 
tractive force to get over obstacles, such as stones. In some 
machines, where it is imj^ortant that there should be very little 
friction at the bearings of axles, the axles are made to lie at 
each end in the angle formed by two wheels with plain rims. 
The main axle rolls on these wheels, and it is only at the axles 
of the wheels that there is rubbing. This rubbing is a very 
slow motion, and as the force of friction is but little increased 
in consequence of the weights of the friction wheels, the energy 
lost in friction may be made very small in this way. Every 
curious student is aware of the way in which rolling takes the 
place of sliding in the ball-bearings of cycles. This kind of 
bearing will probably be greatly used in ordinary machinery. 
The resistance is as if we had a co-efficient of friction inversely 
proportional to the diameter of the ball or friction roller, 
because of indentation of the rolling surfaces. 

If you compare Watt's parallel motion which is still used 
in some pumping engines to cause the piston-rod to move in a 
straight line, with the slide which is now so common, you will 
see that there is very much less loss of energy by friction when 
the parallel motion is employed, because, whereas in the slide 
the rubbing motion is as much as the motion of the piston, in 
the parallel motion rubbing only occurs at the pins of the 
arrangement. Unfortunately, this arrangement does not allow 
the piston-rod end to move exactly in a straight line, and 
produces some friction between the piston and its cylinder, and 
between the piston-rod and stuffing-box ; and it is also much 
more costly and less compact than slides. Hence slides are 
now in general use. 

58. In a journal of length I and diameter ^, if w; is the load and 
^t/w the force of friction, at n turns per minute, the velocity heing 
proportional to nd, the rate per second at which heat is developed 

per square inch of area is proportional to • . Calling - the 

Ldj Id 

pressure p, we see that pnd ougM to he constant in all journals 

if they are to have ahout the same rise of temperature above 

surrounding objects, because the giving out of heat by a surface 

per square inch may be taken as proportional to this rise of 

temperature. This rule is found to be somewhat misleading for 



70 APPLIED MECHANICS. 

labricated bearings, "because fx is not by any means constant. The 
values of pnd found in practice are 1,000,000 to 1,500.000 in 
locomotive crank pins, calculated frr)m full pressure and speed: 
250,000 in marine-engine crank pins: 60.000 to 200,000 in stationary 
engine crank pins : crank sbaft bearings, 36,000 ; railway carriasre 
axles, 300,000. Wben wbat is understood to be a co'nstant in 
one's tbeory varies between 30,000 and 1.500,000, it strains one's 
sense of bumour to maintain the gTavity necessary in the writer of 
a text book. It is evident that we must pay more attention to 
mere pressure ; and it may be said that in practice it is the rule 
not to greatly exceed 200 lbs. per square inch in shafting, unless 
there is bath lubrication, and then the limit is 500 lbs. per square 
inch : 600 lbs. per square inch in crank pins ; IjOOO lbs. per square 
inch in crosshead pins. 

The practical engineer has by processes of success and failure 
arrived at di m ensions in machine design which we have always the 
desire to see reasons for in our theory. It is, however, somerimes 
forgotten that a complete theory must be a very compKcate<i one, 
and attempts to deduce (find reasons for) certain very ^useful rules 
(sometimes impertinently called rules of thumb) fk^m very im- 
perfect theory, do not always succeed. 

It would be interesting to find out why it is the universal 
practice of good engineers to make the ratio of length Z of a 
bearing to its diameter d increase nearly in proportion to the 
nnmber of revolutions per minute. It has usually been lost sight 
of that these bearings never occur in long lengths of shafting — 
only in separate machines like fans, centrifugal pumps, and dynamo 
machines. In long lengths of shafting, "bearings, as their name 
implies, are mainly used as mere supports ; but in separate 
machines they no;t only carry weight — their function is, especially 
in light machines running at great speeds, to keep the shaft fixed 
in direction. If then there is any bending mc'inent ii in the 
spiniile, due to centrifugal force through want of balance, or due 
to other causes, it is easy to show that the pressure per square inch 
of bearing, besides what is due to steady load, is proportional to 
^(l-d, and this multiplied by n is supposed to be kept constant. If >i 
is proportional to the twisting moment, or to the horse-power, or 
d^n, we have a rule Ijd a: n which agrees with the practical one. I 
have myself worked out such a rule for m as a very likely one in 
certain kinds of dynamo machines. In all probability the rule for 
any quick-speed machine would turn out to be this : that if 
questions of cost of construction and space did not intervene, thc- 
ratio of / to d ought not only to increase with n, but also with L, 
the distance between the main bearings of the machine. Where 
there is a possibility of error in the allmeation of the two bearings, 
we have a reason for the ratio of I to d increasing as the square 
root of the speed. 

Exercises . — 1. Find the horse-pjwer necessary to turn a shaft y 
diameter, and making 75 revolutions per minute, if the total load on it 
is 12 tons and <p, the angle of friction, is such that sin. ^ = '015. Ee- 
member that tan. = .u Ans., 2-16. 



APPLIED MECHANICS. 



71 



2. Find the horse -power absorbed in overcoming the friction of a 
foot-step bearing- 4" diameter, the total load being 1| tons, the number 
of revolutions 100 per minute, and the average co- efficient of friction -07. 

Ans., 0-5 nearly. 

3. If it be assumed that the power wasted between the end of a flat 
pivot and its step is proportional at each point to the product of the 
velocity and pressure, what hoi'se-power will be absorbed by such a pivot, 
3" diameter, when running at 120 revolutions per minute, the load on 
the pivot being 2|tons, and the average co-efficient of friction -06 ? Inte- 
gration gives the total energy wasted per second as f a /z w r, where a is 
the angular velocity in radians per second, w is the total load, r is outside 
radius of pivot, and /jt, is the co-efficient of friction. 

Aois., -639. 

4. The length of a journal is 9" and its diameter 6"; it carries a load 
of 3 tons. What horse-power is absorbed in friction when making 100 
revolutions per minute, the average co-efficient of friction being -015? 
What number of thermal units per minute will be conducted away per 
square inch of the brass ? Ans., 0-48 ; 2 centigrade heat units. 

5. A shaft makes 50 revolutions per minute. If the load on the bear- 
ing be 8 tons, and the dia- 
meter of the bearing 7 inches, 
at what rate is heat being 
generated, the average co- 
efficient of friction being 
■05 ? 

If 3 thermal units escape 
per minute when the temper- 
ature of the bearing is 1° C . 
higher than that of surround- 
ing objects, what will be the 
increase in temperature caused 
by the heat produced at the 
bearing? A7is., 19°'6 C. 

59. Friction and 
Speed — You will find 
it instructive to experi- 
ment with such a piece 
of aj)paratus as is re- 
presented in Fig. 28, 
designed to measure the 
friction between sliders 
of different materials 
and the cast-iron sur- 
face p. Here we have 
a pulley with a broad, 
smooth outer surface. 
On this surface lies a 
slide made slightly 
concave, to fit the rim 
of the pulley. On this 

slide we can hang different loads w by the arrangement shown 
in the figure, and- the slide can on^y move a small distance in 




Fig. 27. 



72 



APPLIED MECHANICS, 



any direction on account of stops. There is a fly-wheel to give 
steadiness of motion when the apparatus is worked hy hand. Sup- 
pose, now, that p rotates ia the direction of the arrow. Friction 
causes the slide to move in the direction of motion until it is 
brought up hy a stop. Now let weights he placed in the scalo- 




Pig. 28. 



pan w until the slide is held in a position half way between the 
stops. Evidently the force of friction between the slide and p is 
just balanced by the weight in the scale-pan. With this appar- 
atus you can not only find the co-efficient of friction for two 
rubbing surfaces easily at any speed, but you can very quickly 
vary your experiments, s is a speed counter. 

To what extent I ought to be ashamed of the following facts I 
don't know. They are instructive. Successive generations of 
students at Finshury obtained results from an apparatus like Fig. 28. 



APPLIED MECHANICS. 73 

Jt was arranged to be driven by the college engine at many 
different speeds ; and there was a speed counter. The slide had a 
much longer arc of contact than is shown in the figiu-e, because a 
rocking motion, instead of sliding, was apt to be set up. The load 
was applied at a single point in the centre of the toj) of the slider. 
In every case when the load was kept constant the friction was 
greatest at low speeds ; it got less and less as the speed increased, 
and reached a minimum value ; after which it increased steadily as 
the speed increased, the rate of increase of friction with speed 
getting less towards our highest speeds. I thought these cui'ves 
obtained by students well worthy of special study, and several 
times projected an investigation of my own, and ru-ged others to 
take it up ; but I was too busy with other matters to give it much 
attention. Now, from Prof. Osborne Reynolds's explanation of 
the results of Mr. Beauchamp Tower's experiments one sees how 
very different the phenomenon is from what occurs between flat 
surfaces. Air is being pumped by friction into the space between 
the slider and pulley, and the pressure underneath the slider is 
greater than the atmospheric pressure, and varies from point to 
point, as we have proved by inserting little pressure-gauges. 

We are about to use now another piece of apparatus (Fig. 29), 
the slider being flat, and lying on a circular, flat horizontal 
plate, which may be kept rotating at any speed. The slider is 
prevented from moving much by stops, and friction is balanced, 
as in the above case, by a scale-pan. The apparatus is only 
now (July, 1896) being run for the first time. What sorts of results 
will be obtained from it I do not know. 

60. Although, on the whole, in any machine the average forces 
of friction do not seem to depend much upon speed, and they increase 
in proportion to the load, and in other ways seem to follow the 
laws set down in Art. 55, when we make experiments on the 
friction at any one place in a machine we obtain inconsistent 
results. So long as our theory of an action is wrong, our experi- 
ments give rise to what are called inconsistent results. On the 
hypothesis of Art. 55 the mathematicians have built a science, and 
thousands of examples and exercises have been invented to illus- 
trate it. The exercises and examples are valuable to the engineer ; 
but he must remember that they have been invented by mathema- 
ticians for the training of mathematicians, and he must exercise 
caution in using the results. I give some examples in Art. 58. 
Two substances, when they really touch, get welded together ; and 
.Ms seizing seems to occirr in some journals and footsteps under 
heavy loads. Bodies said to touch or rub on one another are 
really separated by a layer of air or other fluid. A slider like c 
(Fig. 24) is separated from the table a b by a layer of air ; and 
the greater the load, the less is the thickness of air. 

Students will find it very interesting to study the fiiction 
between two scraped surfaces in the workshop. If one plate is 
laid down on the other, there is usually very little friction, because 
there is a thick layer of air separating the siu-faces. By 23utting 
on a load, and gi^dng small sliding motions, we can make 
the layer of aii' very thin-so thin, indeed, that when the top 



74 



APPLIED MECHANICS. 



plate is lifted the bottom one sticks to it, and lifts also, partly 
because there is a partial vacvium between them, partly also, prob- 
ably, because of molecular attraction seeing that it occurs even in 
a good vacuum. Isow, when, through the one plate having lain 
on the other a considerable time, or through pressure, we get the 
layer of air very thin, it is found that there is considerable resistance 
to sliding. In fact in this case, where we might expect to find the 
phenomena of friction assuming their simplest form, we find what 
seem to be the most inconsistent results. When c (Fig. 24) slides, 
it seems as if fi'esh air were being carried into the space between 
the surfaces, keeping thorn apart, and that the greater the velocity 




Fig. 29, 



of rubbing the more air is carried m, so that the separatior. 
between the surfaces is proportional to the velocity of rubbing. 
When we come to discuss fluid friction, and reflect that all the 
friction we know of is really fluid friction, we shall not be aston- 
ished that the laboratory results from the rubbing of solids are 
sometimes inconsistent -looking — we shall wonder greatly that the 
above-mentioned law should be even aj)proximately true. But I 
cannot think the eifect merely one of the fluid, there is also 
molecular attraction. In any lubricated bearing which is not kept 
flooded with oil, the rules called the laws of solid fi'iction are found 
to be approximately true — that is, we may take the load on the 
bearing in pounds, multiplied by a coeificient yu, as representing a 
force of friction ; and this, multiplied by the distance of rubbing 
in feet, is the mechanical energy converted into he-at by friction. 
fx is less as the temperature is gTeater, partly because the viscosity 



APPLIED MECITANICS. 75 

of a fluid, diminishes with temijeratiire (see Art. 63) ; but it is not 
only because of this, for the body of the lubricant also alters with 
temperature — that is, it tends more to get squeezed out of place. 
Of course the friction dej)ends greatly upon the nature of the 
lubricant, and. the phenomena are really so very complicated that 
in the ]present state of our knowledge the reader must perforce be 
satisfied with the rough general law that I have mentioned. That 
the supply of oil, however small, shall be continuous, and not inter- 
mittent, is regarded as the most important condition in the lubrica- 
tion of bearings. The lubricant ought to suit the nature of the 
load. Thus great body is necessary when the loads are great, so 
that the oil may not be squeezed out; and greases and solid 
lubricants, such as soapstone and j)lumbago, must be used for very. 
hea^^ loads. It is also usual to cast plugs of white metal and 
other soft alloys in recesses of the step, and in some cases to line 
the whole step with such a soft alloy. As for the detailed 
construction of pedestals, hangers, a frames, and other supports for 
shafting, this is to be learnt in the drawing-office and shops, and it 
would be useless to refer to it here. 

61. As it has been found that with some kinds of material the 
statical friction — that is, the friction which resists motion 
from rest — is someivhat greater than the friction of the surfaces 
when actually moving, experiments have been made to deter- 
mine whether, at very small velocities indeed, with such 
materials, there is not a gradual increase in the friction. It is 
known that at ordinary velocities the friction is much the 
same as at a velocity of -01 foot per second. We have reason 
to believe that with metals on metals and air between, there is 
the same friction at all velocities, even down to one-five- 
thousandth of a foot per second^ whereas with metals on wood 
the friction increases gradually as the velocity diminishes, until 
when the velocity is 0, the friction is what we call static 
friction. Again, at very high velocities it has been found that 
there is a very decided diminution of tlie coefficient of friction 
between a cast iron railvmy brake and the wrought iron tyre 
of a wheel. The coefficient was '33 for very slow motion, -19 
for a speed of 29 feet per second, and -127 for a speed of 66 
feet per second. It has also been observed in these railway 
brake experiments that when a certain pressure is applied for 
a short space of time the friction diminishes. All such results 
as these, however interesting they may be to the railway 
engineer, tell us nothing about what I have hitherto called 
friction, because I have supposed the rubbing surfaces to 
remain unaltered, whereas these railway brakes are rapidly 
worn away, and the effects of abrasion and polishing are of an 



76 APPLIED MECHA>'ICS. 

utterly different kind from the effects of friction of which I 
have hitherto been speaking. 

62. We must remember that although friction leads to wast« 
of energy, all the enerory spent in overcoming friction being 
converted into another fonn of energy called heat, still the 
force of friction is very useful. The weight resting on the 
driving-wheels of a locomotive engine multiplied by the co- 
efficient of fiiction between the wheels and rails represents the 
greatest pull which the engine can exert upon a train. Suppose 
the weight on the driving-wheels to be 15 tons, and that the co- 
efficient of friction of wrought iron on wrought iron is about 2, 
the greatest pull which the locomotive can exert is 15 x 0'2, or 
3 tons. If the train, including the locomotive itself, resists witli a 
greater force than this, the driving-wheels must slip : if the train 
resists with a less force than this, there is no slipping, the wheels 
simply roll on the rails. Again, it is the friction between 
the soles of our feet and the ground that enables us to walk ; 
friction enables us to handle objects ; fiiction enables a nail to 
remain in wood : friction keeps mountains from rolling down. 

63. Fluid Friction. — I have been considering the friction 
between solid bodies only. The friction between liquids and 
solids or between liquids and liquids is of a very different kind. 
If a man attempts to dive into water unskilfully, and falls prone, 
you know that the water ofiers a very considerable resistance 
to a change of shape. Xow this is mainly the resistance that 
any body offers to being rapidly set in motion. If you came 
coUi'iing against the end of the most frictioniess carriage, you 
would also experience its resistance to being suddenly set in 
motion; whereas the constant steady resistance to motion 
which the carriage experiences when moving with a uniform 
velocity is called friction. What I wish rather to refer to is 
the resistance to the motion of water in a pipe, the resistance 
to the steady motion of a ship. 

In nearly all ordinary cases the motion is complicated and 
difficult to study. The simplest motion is in plane parallel 
layers. Imagine two infinite plane parallel boundaries with 
the fluid between : one of the boundaries at rest, the other 
moving with uniform velocity v in its own plane. Imagine 
the fluid to stick to each boundary. If 6 is the distance 
between them, the tangential force per unit area required to 
keep up the motion is ^ v -f- 6 if /i is the coefficient of 
viscosity. Theory shows that ^ ought to be constant if the 



APPLIED MECHANICS. 



77 




motion is truly in plane layers. As we cannot experiment 
with infinite surfaces, I thought that I could approach the 
condition most nearly with the apparatus shown in Fig. 30. 
F is a hollow cylindric body supported so that it cannot move 
sidewise, and yet so that 
its only resistance to 
turning is due to the 
twist it would give the 
suspension wire, A. c c is 
water or other liquid fill- 
ing the annular space be- 
tween the cylindric sur- 
faces D D and E E, and 
wetting both sides of F. 
When the vessel d d, e e 
is rotated, the water mov- 
ing past the surfaces of P 
tends to make F turn 
round, and this frictional 
torque is resisted by the 

twist which is given to the wire. The amount of twist in tlie 
wire gives us, then, a measurement of the viscosity of liquids, 
and investigations may be made under very different conditions. 
The above apparatus was designed and partly constructed 
in Japan in 1876. Experiments made with it by Finsbury 
students on olive oil are described in the Proceedings of the 
Physical Society of London, March, 1893. At constant 
temperature below a certain critical speed, I found that the 
friction was proportional to the velocity, so that /j. could be 
found. At that critical speed I found that there was a sudden 
change in the law, and above that speed the friction is pro- 
portional to a higher power of the speed than 1. We know 
that above the critical speed the plane motion which I 
described above would become unstable, and eddies would be 
formed. From a theoretical point of view it is curious* that 

* Some experiments of Mr. D. Baxandall, not yet published, show that 
the forces of friction— that is, the resultant forces applied to the solid bodies 
which form the boundaries of a mass of fluid to maintain relative motion — are 
strictly proportional to the relative, velocity at small speeds, and tliere is 
always a critical speed above which the friction is proportional to a higher 
power of the velocity. We have tried surfaces arranged like those of churns, 
with paddles and curiously shaped vanes, and the law is always true. With 
mixtures of glycerine and water, the higher power of the velocity above 
referred to depends on the proportions of glycerine and water. 



78 



APPLIED MECHANICS. 



the phenomenon should have been so marked between my 
cylindric surfaces, because even at slow speeds cylindric motion 
ought to be unstable inside a fixed cylindric surface — that is, in 
the inner part of my trough. 

At speeds below the critical, I measured /z at many 
different temperatures, and noted the rapid decrease in it as 
the temperature increased. 

Yery interesting observations may be made at small speeds 
by immersing similar and equal heavy discs of brass in air, 
water, and oil, suspending them by fine steel wires. (See Fig. 30.) 
When the suspension wires are twisted and let go, the bodies 
vibrate like the balance of a watch. But it is only the one which 
vibrates in air that goes on vibrating for a long time ; the one 
in water keeps up its motion longer, however, than the one in 
oil, showing that there is more frictional resistance in oil than 
in water, and more in water than in air. The rates of diminu- 
tion of swing or the stilling of the mbrations tell us the relative 
viscosities of the fluids. If, by means of a pointer or minor 
attached to the wire, you observe the various angular displace- 
ments, noting the time for each, and then plot your observa- 
tions on squared paper (as in Art. 53), you will find what is 
very nearly a curve of sines for the vibrations in air ; and for 
the different liquids damping curves, which show the effect of 
friction in the liquids. Similarly, the rate of diminution of 
swing of the vibrating fluids in U tubes, one containing water 
and the other oil, tells us about the relative co-eflicients of 
viscosity of the liquids. 

64. The motions in these cases are not so simple as in the case 
which I considered experimentally. The c^uestion of the resist- 
ance to the passage of fluids through pipes is one which has 
attracted much attention, and the results of experiments seemed 
very inconsistent until Professor Osborne Keynolds considered 
the problem. It was known that the pressure difference at the 
ends of a level uniform pipe necessary to produce a certain 
flow was proportional to the length of the pipe, and it was usual 
to say that the force of friction was proportional, as in all 
other cases of fluid friction, to the wetted area ; it is quite 
independent of the pressure ; it is pro])ortional to the velocity 
of the water when the velocity is small, but at high speeds it 
increases much more quickly than the speed. Thus, as I said 
in the first edition of this book, of water flowing in a certain 
pipe, " at the velocities of 1,2, 3, etc. inches per second, the 



APPLIED MECHANICS. 



79 



friction is proportional to the numbers 1, 2, 3, etc., whereas 
at the velocities of 1, 2, 3 yards per second the friction is pro- 
portional to the numbers 1, 4, 9, etc." At small velocities, 
three times the speed means three times the friction ; whereas 
at great velocities, such as those of ships, three times the speed 
means nine or more times the friction. We see, then, that 
friction in fluids is proportional to the speed when the speed is 
small, to the square of the speed when the speed is greater, and 
at still greater speeds the friction increases more rapidly than 
the square of the speed. The resistance to motion of a rifle 
bullet is proportional to the square root of the fifth jDower 
of the speed ; that is, a bullet going at four times the 
velocity meets with thirty -two times the frictional resistance 
from the atmosphere. (See Art. 68.) Again, it has been 
found that the friction is much the same whatever be the 
pressure. Thus it is found that when the disc and liquid 
apparatus is placed in a partial vacuum or under considerable 
pressure, there is exactly the same stilling of the vibrations. 

This fact is illustrated by the apparatus, Fig. 31. Water 
tends to flow from vessel A to vessel b, through the long tube. 
Whether the tube is in the position shown in Fig. 31, or in 
the position Fig. 32, or is acting as a syphon, we find the same 
flow through it ; the same quantity of water passes through it 
per second, although the pressure of the water in the tube in 




Fig. 31. 



the position Fig. 32 is very much 

greater than in the position Fig. 31, 

or again when the tube is a syphon. 

In the apparatus actually used by Fio.. 32. 

me, there is a stopcock in the 

middle of the tube, and by nearly closing it one is sure 

that the friction occurs at the place where the pressure is 




80 



APPLIED MECHANIC! 



greiitest in Fig. 32. The comparison is most readily made 
by observing how long it takes for a certain change of levels 
to take place in the two vessels, repeating this several times 
with the tube in various positions, beginning and ending each 
experiment with the same difference of levels. Again, fluid 
friction, for even considerable velocities, does not seem to 
depend much on the roughness of the solid boundary. This 
seems to be due to the fact that a layer of fluid adheres to 
the solid surface and moves with it. Even when the disc of 
Art. 63 is indented, or when large grooves are cut in it we 
find practically the same frictional resistance. 

Comparison of the Laics of Fluid and Solid Friction. 



Friction between Solids. 



Fluid Friction. 



1. The force of friction does 
not much depend on the velocity, 
but is certainly greatest at slow 
speeds. 

2. The force of friction is pro- 
portional to the total pressure 
between two surfaces. 

3. The force of fi-iction is in- 
dependent of the areas of the 
rubbing surfaces. 

4. The force of friction depends 
very much on the nature of the 
rubbing surfaces, their roughness, 
etc 



1. The force of friction very 

mucb depends on the velocity, and 
is indefinitely small when the speed 
is very slow, 

2. The force of friction does not 
depend on the pressure. 

3. The force of friction is pro- 
portional to the area of the wetted 
surface. 

•4. The force of friction at 
moderate speeds does not much 
depend on the nature of the wetted 

surfaces. 



65. Molecular theory gives us the cause of fluid fiiction in such 
a fluid as air. Layers of fluid at different velocities are continually 
interchanging molecules by ordinary diffusion : consequently, the 
relative motion is being destroyed, the rate of loss of momentum 
by one layer and gain of it by the other enabling us to state that 
the force required to maintain the motion is proportional to the 
surface of contact and to the relative velocity. In regard to 
friction in gases, the explanation is complete, the greater diffnsivity 
at higher temperatures causing the viscosity to be greater also. 
Indeed, viscosity is proportional to the square of the absolute tem- 
perature. In the same way, if two trains were passing one another and 
the same number of passengei-s jiunped from each train to the other, 
the trains would become more equal in speed ; there would seem 
to be a mutual frictiona' force between them proportional to the 



APPLIED MECHANICS3. 81 

rate of loss or gain of momentum per second. But in liquids there 
is less viscosity at higher temperatures, although there is greater 
diifusivity. This is probably diie to the fact that mere diffusivity 
is all-important in gases, the molecules of which exert no forces 
upon one another except by collision ; whereas in liquids, although 
tlie greater diffusivity at higher temperatures would tend to make 
them behave like gases, in regard to viscosity, forces are always 
acting between the molecules which resist shearing strain, and 
these forces get less as the temperature increases. 

Now, in any case of relative motion between the boimding 
surfaces of a fluid, beyond a certain velocity, motion in plane 
layers becomes unstable and sinuous motion sets in. This means 
that the surfaces across which interchange of momentum by 
diffusion may take place become greater in area ; so that above 
a certain critical speed I take it that we may expect almost any 
law connecting friction and speed. Theory shows that for any 
given shape of surface the critical speed will be less as the density 
of the fluid is greater, and it is less as fj. is less. A very friction- 
less fliiid is very unstable. 

I believe that all friction said to be between solid surfaces is to 
be regarded as taking place in the fluid which always separates 
such surfaces. This statement seems a mere truism. It is like 
many another yet to be made by discoverers in applied physics. 
As a matter of fact, the above table showing the utter difference 
in character between the phenomena of solid and fluid friction 
quite hid from everybody's view the fact that all friction must be a 
fluid friction, until Professor 0. Reynolds opened our eyes. He has 
gixeii us in his lectures at the Eoyal Institution and in his paper 
published in the Transactions of the Royal Society the suggestion 
that it is to some extent in the solution of hydrodynamic problems 
we must look for an explanation of the curious phenomena of solid 
friction. I have already mentioned a curious phenomenon often 
brought to my notice in connection with the use of the apparatus 
shown in Fig. 27. Let mo now describe some experiments made 
on the friction of journals. Probably everybody has been occa- 
sionally interested in curious results obtained when testing oils with 
the Thurston oil-tester (Fig. 33). Many of these will be found 
published in Mr. Thurston's book on " The Materials of Engi- 
neering," Part I. ; but every mechanical laboratory ought to 
be provided with the apparatus, that students may study the 
phenomena for themselves. In Hirn's experiments, made in 1855, 
he found that the force of friction was proportional to the square 
root of the product of load and velocity. In the experiments of 
Mr. Beauchamp Tower upon a steel journal with a gun-metal cap 
onlj. the cap being loaded, the lubrication being practically an 
oil bath, the friction was found to be practically independent of the 
load for loads so excessive as from 100 to 520 lbs. to the square 
inch (the diameter being 4 inches and length 6 inches, the 
pressure is taken as the whole load divided by 24), and in all cases 
to be practically proportional to the square root of the velocity. 
If, instead of such excessive lubrication as we have in an oil bath, 
there was only the lubrication due to an oily pad pressed ngaicst 



APPLIED MECHANICS. 




Fig. 83. 

the journal below, the ordinary law assumed for the friction of 
solids was found to he approximately followed. In collar hearings, 
as in the thrust hearings of a propeller shaft, he again found that 
the ordinary laws of solid friction are fairly well followed, and 
that only very much less pressures (75 Ihs. per square inch at high 
speeds and 90 Ihs. per square inch at low speeds) were possible 
without seizing. These curious phenomena have been completely 
explained hy Professor 0. Reynolds. They depend upon the 



APPLIED MECHANICS. 83 

iarrj-ing- of the lubricant into the space between the step and the 
journal, and if there is not an oil bath and the motion is all 
Ln one direction, the lubricant leaves the place -^here it is most 
wanted and the journal seizes at comparatirely low pressures ; 
whereas if there is such an irregularity of motion or reversal of 
motion as helj)s the lubricant to maintain its place, very gToat 
pressures may be employed. Thus in crank pins and railway 
axles pressures as high as 400 lbs. per square inch with sperm oil 
and 600 lbs. with mineral grease ha^-e been used ; and, indeed, in 
slow-mo^dng steam engines nearly double these pressiu-es have 
been used. It is well to remember that at the place where the 
journal most nearly approaches the step the pressure in the oil 
becomes very great, and if there is an opening there the oil is 
forced out. It is now quite common to employ a force-pump to 
pump oil at great pressure into these parts of the bearings, and in 
consequence much higher loads on bearings are possible than used 
to be the case. 

In any case, we must tiy to understand the distribution of 
pressure in the bearing, so that in our endeavours to utilise an 
ordinary syphon-lubricator for example, we shall not be attempting 
impossible things. 

66. It is in the di^a wing-office and shops that students wiU become 
acquainted with the methods in actual use for supporting horizontal 
shafts. Footsteps for vertical shafts, which give endless trouble in 
many high factories, are now in many cases water- or oil-borne, 
being converted into the rams of hydraulic presses having only 
a very small range of vertical motion, or, rather, so arranged that 
the lifting force of the fluid shall always be less than the weight of 
the shaft. 

67. It is when great forces have to be overcome slowly, and 
particularly mth a long translational motion, that water-pressure 
machinery shows itself most greatly superior to other machinery, 
for friction seems to be nearly independent of pressure. But if in 
any place the water is set in rapid motion there is internal friction 
and waste of energy. Where fluids move so slowly that the 
friction is proportional to the velocity, we seldom consider it in 
our engineering work. At the valves of pumps and in pipes it 
is usually, on the whole, economical to let energy be wasted in 
water friction. There is a certain relationship between velocity, v, 
and diameter, d, of pipe for a particular flxud which causes a certain 
V to be critical. Below that value of v the water flows in straight 
streams ; at that critical value the beautiful straight lines which 
Professor Reynolds shows coloured with aniline dye suddenly break 
up into confused, smoke-like eddying cloud. Below the critical 
velocity the total pressure difference — or friction, as we may call 
it — ^required to keep up the flow is proportional to the velocity. 
Above that critical velocity the friction is proportional to a power 
of the velocity which varies from 1-7 to 2, depending upon the 
nature of the material of the pipe. 

68. The mathematical investigation of the resistance to the pas- 
sage of a body through a viscous fluid is so difficult that we have 
almost no results which may be relied upon. Without viscosity 



84 APPLIED MECHANICS. 

there would be no resistance to steady motion, whatever tho shape of 
the object. It is difficult to imagine that there would be no propelling 
force on a sailing-boat if the air were fiictionless, and yet this is so. 
Even in the case of a ship, experiments on which have been going 
on continuously since ships were first built, our knowledge is very 
incomplete. Roughly, we may take it that resistance is generally 
proportional to square of speed. In the case of shot this law holds, 
probably up to speeds of 300 feet per second ; fiom 400 to 1,000 
feet per second the resistance is possibly proportional to the 2^ 
power of the speed. Beyond 1,100 feet per second we may take, 
F being in pounds, d the diameter of a shot in feet, v the velocity 
in feet per second, f ■=-fdr- [y — 800), where /= 3 for spherical 
and 2 for elongated shots with ogee-sLiped heads. The velocity is 
greater than that of sound, and probably it is to this that the 
change of law is due. The fact that even in the steadiest winds 
there is pulsation, causes scientific speculation about wind pressure 
to be difficult. 

69. Eeynolds has deduced from hydrodynamics the rational 
formula 

lb" p2-n d'i-S v^/a . . . . (1) 

as the loss of energy per pound of fluid passing through a 
pipe of length l feet and diameter d feet at v feet per second, 
(I have reduced his numbers to suit the foot as the unit of length.) 
The index n\s,\ for velocities below the critical velocity, v^ =r -039 
p/d, and n varies from IT to 2 at higher velocities than the critical, 
r is proportional to the co-efficient of viscosity, which changes 
with temperature. In the case of water he takes 

P :3 1 -^ (1 _ -0336 Q + -000221 a^) . . . . (2), 
where 9 is temperature Centigrade. 

A= 1-917 X 10*-'; - B=:36-8. 

Note that the critical velocity depends upon the temperature 
and size of pipe. Thus, for a tube -jLth of an inch in diameter, the 
critical velocity is 4-65 feet per second ; for a pipe 1 inch in 
diameter the critical velocity is -465 feet per second ; for a 6-inch 
pipe {d ziz 05) the critical velocity is "077 feet per second. In all 
practical hydraulic cases the critical velocity is exceeded, and for a 
general rule, with cast-iron pipes in actual use, we usually take 
n = 2. In this case, in (1), the influence of p, the temperature 
term, is unfelt ; that is, in practical hydraulic work, temperature 
has no important influence. The fomiula now becomes 

LB-y^AD or -0007 lv2/d . . . . (3). 

As a mnemonic for this simple formula, let the student imagine that 
a solid pi-ism of water of length l is moved along a pipe rubbing 
all round its perimeter, the friction being proportional to the square 
of the velocity and to the area of the rubbing surface. Thus, if * 
is the wetted perimeter, ls is the area and the force of friction is Lsr^ ; 
that is, if ^ is the pressure difference which produces the motion 
and A is the area of cross-section, p\ ocj^sv^. The loss of energy per 

pound being proportional to p, this oc lv^ — . Now, a/« is caUed 



APPLIED MECHANICS. 85 

the hydraulic mean depth, m, of any channel, and we find loss of 
energy per pound = chv^/tn .... (4). 

In the case of a round channel full of water 

TT 9 _ 1 

m=r-D^ -r 7rD=— D. 
4 4 

Comparmg (4) with (3), the formula of Osborne Reynolds corre- 
sponds to c = -000175. 

The constant in the formula (3) agTees with that of D'Arcy 
for small pipes. Reynolds has compared (1) with D'Arcy's experi- 
ments as well as with his own, from the smallest sizes of pipes to 
20 inches diameter, and finds that there is practical agTce- 
ment. D'Arcy's pipes had joints which somewhat satiated the 
resiilts. Reynolds gives for n the values : — Lead- jointed pipes, 
1'79; varnished, 1*82; glass, 1*79; new cast-iron, 1-88; incrusted 
pipe, 2-0 ; cleaned pipe, 1'91. This formula of Reynolds is rational, 
and suits every imaginable size of pipe, and I prefer it. But that 
of D'Arcy is more commonly used for pipes of from 3 inches to 
2 feet in diameter, and it is often used in academic exercises, in 
which, indeed, almost any loss of energy per pound of water 
(usually called loss of h^ad) is expressed as / x the kinetic energy 

per pound of water, or / x h~ • D'Arcy gives 

/ = -02 (l -f-i. ^ 1 
for a straight pipe of length l feet and diameter d feet. 

70. Resistance to Rolling. — When one wheel or cylindric 
body rolls upon another there is some conversion of mechanical 
energy into heat. The power lost seems, roughly, to be propor- 
tional to the force pressing the two bodies together, to the 
velocity of rolling, and to the curvature of the smaller of the 
two. We have very little experimental knowledge of the subject. 
In all probability the power wasted is proportional to the 
strain energy per second stored in the material, the waste 
being due to viscosity. When the velocity is very great, as at 
the driving-wheels of locomotives, secondary effects are pro- 
duced, waves of compression and extension travelling in the 
rim of the w^heel and in the rail, with very curious re-sults. In 
some experiments which I have made with great pressures 
between hard cast iron wheels rolling upon one anothei, there 
seems to have been much local heating just at the surfaces. 
At the end of. some months of work a quantity of black dust 
had been produced, and each particle, when examined by the 
microscope, looked like a piece of slag. 

Besides energy wasted by changing strain in the material, 
there is slipping at the surfaces in contact. A student who 



Fig. 34. 



86 APPLIED MECHANICS. 

remembers that when a strut is compressed it swells, and when 
a tie bar is lengthened it gets thinner, can study the " creep " 
which occurs both here and in belting, for himself. Imagine 
points one inch apart upon the rim of an iron wheel, and 
another set upon an unstrained plane indiarubber surface. 
Now draw the wheel as it indents the surface. As in Fig. 34, 

points 1, 2, 3, and 4 
1 are further apart, and 

I points 6, 7, 8, 9 are 

} nearer together than 

I in the unstrained con- 

} j^ r dition, and hence the 

I ^ ^ y metal and indiarnb- 

i 1^<^rT~"'"-^^ ^^^^ surfaces slide 

J A. J;^^^ ^ — 1- -- --- ■■ npon one another. No 

ordinary material 

coating seems to have 
much effect in pre- 
venting the sliding. A cast iron wheel on planes of cast 
iron, boxwood, and on indiarubber, seems to have frictional 
resistances to rolling in the proportion of 1 : 2 : 8. 

The loss of energy in belting is partly due to this, partly 
due to energy wasted in bending and unbending the belt. 
Both the lessened distance of rolling and the slip of a belt 
seem to be proportional to the power transmitted. M. EafFard 
has actually used a dynamometer on this principle. He 
transmits his power, to be measured, by means of a thick 
indiarubber belt through two equal pulleys; the difference 
of speed of these pulleys is taken to be a measure of the 
power. The slip is quite noticeable when speed cones are used 
in driving machines at various speeds with variable power, for 
the actual speed has to be carefully measured ; calculation 
from the known sizes of the steps of the cones giving in- 
accurate results. 

When pressures are not too great, as in the ball bear- 
ings of cycles and some machine tools, there can be no doubt 
whatever of the ease of running. Fig. 35 shows the ordinary 
adjustable ball bearing used in bicycles, d is the fork and h the 
hub of a wheel. The spindle A is fixed to the fork d. Oue of 
the hard steel cones c is tight against a shoulder v ; the other 
c' is tightened just enough to let the wheel revolve easily, and 
then it is locked by the lock-nut K. The linings of the shaped 



APPLIED MECHANICS. 



87 



ends of c are hardened steel, and a number of hard steel balls 
are placed between. Fig. 36 is an enlarged drawing of the 




Fig. 35. 



Fig. 36. 



ball and the linings, showing that the radii of curvature of the 
ball, cone, and cup are different; the friction of the bearing 
will be much less than if the radii of curvature were nearly 
the same. A very little oil getting inside the hub finds its 
way to the balls. 

Some experiments [Proc. I. C. E., Yol. 119, p. 456) on rollers 
between flat cast-iron plates, give as the resistance in pounds to 
rolling c/ v^r where r is radius in inches and c=-0063 for cast-iron, 
•0120 for wrought iron, 0073 for steel. These are 13 per cent. 
gTeater for wrought iron plates and 13 per cent, less for steel plates. 
The crushing load in pounds on a wrought iron roller seems to be 
444 r per inch of its length. 

Several experimenters are now engaged in procuring for us 
more exact information on rolling friction. 

In using roller hearings on carriages, it has heen found that 
there is a duninution of from 23 (on gTadients of 1 in 20) to 60 per 
cent, (on gradients of 1 in 140) of the tractive effort required with 
ordiiiary hearings. In ordinary machinery, the loss of energy by 
friction has heen found (in one experiment) to he less than one- 
thii-d of what it is with good ordinary hearings. Oil is only needed 
to pre^-ent rusting. 



S)i 



CHAPTER Y. 



EFFICIENCY. 

71. Mechanical Advantage. — In books on mechanics you 
will usually find that when simple machines are described, they 
are only considered in relation to their 
Mechanical Advantage. That is, suppose a 
small weight e, now usually called the 
effort, is able by means of the mechanism 
to cause a larger weight, r, usually called 
the resistance, to rise, the ratio of r to e 
is called the mechanical advantage. Now, 
in nearly all cases you will find that, 
when there is a mathematical investigation 
of a machine, the assumption is made that 
there is no friction. I have already 
shown you that the problem of 
taking friction into account is a 
very difficult one. But, as we have 
seen, a practical man can experi- 
ment on the effect of 
friction; and, happily 
for us, he obtains re- 
sults which are gener- 
ally very simple. Let 
the reader make a few 
experiments himself, 
or let him by means of 
squared paper find the 
relation between e and 
II from the following 
results, taken from a 
crane. Fig. 37, whose 
gearing was well oiled, 
jr^g. 37^ and whose handle was 

replaced by a grooved 
wheel, round which was a cord supporting e : — 




APPLIED MECHANICS. 89 



R. 






E. 


esislance just Effort just able to 


Overcome. Overcome Resistance. 


100 lbs 8-5 lbs. 


200 „ 






12-8 „ 


300 „ 






17-0 „ 


400 „ 






21-4 „ 


500 „ 






25-6 „ 


600 „ 






29-9 „ 


700 „ 






34-2 „ 


800 „ 






38-5 „ 



We found that e fell forty times as rapidily as R rose, and 
you may have imagined that the mechanical advantage was 
forty, or that a weight, e, could lift a weight, R, forty times as 
great as itself. This would be true if there were no friction ; 
but we see that in practice it is not the case. Plot the above 
values of e and R on squared paper, and you will find that, if 
the weight r is increased 1 lb., e must be increased '0429 lb. ; 
and also that when r is 0, an effort e of 4*21 lbs. is needed to 
cause a slow motion of the crane ; so that the law is 

Er= 4-21 + -0429 R. 
N^amely, multiply the resistance r in pounds by the fraction 
•0429, and add 4-21 : the answer is the effort required to lift 
R. When you have worked out this rule, employ it in finding 
how much effort, e, is required to lift a ton with such a crane. 
— A?iswer, 100 "3 lbs. 

The word power is generally, but very unscientifically, used 
as the name of the force which I have called e, the effort. 
Power may, of course, be used in many senses by newspapei 
writers, but when used by the engineer it is a technical term, 
meaning the rate of doing work. If a weight of 1,000 lbs. 
falls 100 feet in two minutes, it does 1,000 x 100 or 100,000 
foot-pounds of work in two minutes, or 50,000 foot-pounds of 
work in one minute. Now, 33,000 foot-pounds of work done 
in one minute is called a horse-power, and hence our falling 
weight gives out 50,000^33,000 or 1-5 horse-power. Ten 
horse-power means ten times 33,000 foot-pounds of work done 
in one minute. The idea, then, of power is an idea of work 
done in a certain time. 

72. Economical Efficiency. — Take any pair of numbers from 
the above table, say e = 8-5 lbs., when r = 100 lbs. Let us 
suppose that E is moving at the rate of forty feet per second, 
then we know that R is rising at the rate of one foot per 



90 APPLIED MECHANICS. 

second. E is giving out the power 8 5 x 40^ or 340 foot-pounds 
per second ; r is receiving 100 foot-pounds per second. The 
ratio of the power usefully employed to the power given to the 
machine is called the efficiency of the machine, so that our 
crane has an efficiency 100^340, or -294. Sometimes the 
efficiency is put in the form of a fraction ; sometimes we say 
that it is 29 '4 per cent., meaning that the machine employs 
usefully 29-4 per cent, of the energy given to it. 

Now take another pair of numbers, say e = 38 "S, r = 800, 
and let e fall forty feet in one second, as before. We 
now get as our answer -519 — that is, more than half, or 
51 "9 per cent., of the power given to the crane is usefully 
employed. AYe see, then, that as the power given to the crane 
is gTeater, the efficiency is also greater. This arises from the 
fact that the friction of the unloaded crane is always entering 
into the calculation ; and if we take the case where no resist- 
ance, R, i^ being overcome, and e must be 4 -21 lbs., we shall 
find an efficiency, 0, because work is being given to the crane, 
and none is coming out usefully. You will always find that the 
power usefully given out is a certain fixed fraction of the total 
power given to the machine, minus the power required to drive 
the crane at the given speed when it is unloaded. Choose some 
speed, say that E falls forty feet per second ; find the total 
power or ^0 E ; find the usefully employed powder 1 x R for 
every case of the above table. Plot your answers on squared 
paper, and you will find this rule : if Pq is the power required 
to drive the crane at the same speed when unloaded, if u is 
the useful power, and t is the total power supplied : — 

T = l-761 u + Pq 

73. In some machines there have been attempts to ap- 
portion the loss of power to various parts. As a rule, these 
are very speculative. Roughly, we may say thao the frictional 
loss in a steam-engine may be divided in the folloAving propor- 
tions : — Crank-shaft bearings and eccentric sleeves, 1 ; valve, 
if unbalanced, -6 ; valve, if balanced, -05 ; piston and rod, -4 ; 
crosshead and slides, "2 ; crank pin, -14 ; total loss because of 
air pump, 0-3 to 0-5. It is fairly obvious that on account of 
the great weight of the fiy-wheel and other parts much of the 
energy loss in a steam-engine is the same w^hether the engine 
is giving out much or little power, and in many cases for 
purposes of calculation tliis assumption may be made. 



APPLIED MECHANICS. 



91 



74. The work of this and the succeeding two pages is more 
suggestive than any other work in this book. From the follow- 
ing results of experiment with a gas-engine, show, by plotting 
on squared paper and correcting for errors of observation, that 
if I is the indicated horse-power, b the brake horse-power, g the 
cubic feet of gas per hour, including what is used for ignition, then 
G = 20-3 I + 8, G = 20-4 B + 45, B = 1—1-8. 



I 


B 


G 


G/I 


g/b 


Efficiency. 


13-4 


11-6 


280 


20-9 


24-1 


•166 


10-2 


8-4 


216 


21-2 


25-7 


•155 


7-3 


5-4 


156 


21-4 


28-9 


•123 


4-6 


2-9 


104 


22-6 


37-9 


•112 


1-8 





45 


25-0 


— 






When we plot the values of g and i and of g and b on 
squared paper, we find points lying very nearly in straight 
lines. Assuming that fchey ought to lie in straight lines, we 
find the above laws satisfied. 

Let the student fill in the columns showing g/i and g/b. 
Also from the brake horse-power, and knowing that one cubic 
foot of gas per hour means an actual supply of energy of a 
quarter horse-power, let him fill in the column of efficiencies. 
[The calorific power of one cubic foot of average coal gas may 
be taken to be from 500,000 to 540,000 foot-pounds; of 
Dowson gas, it is about 124,000 foot-pounds.] 

Again, taking the following experimental results from an 
oil-engine (one pound of oil being taken to give out 11,700 
centigrade heat units in burning), i being the indicated, b the 
brake horse-power, and o the pounds of oil used per hour. 



I 


B 





O/I 


O/B 


Efficiency. 


7-41 


6-77 


6-4 


•86 


•95 


•128 


8-33 


6-88 


6-8 


•82 


•99 


•122 


4-71 


3-62 


5-0 


1-06 


1-38 


•088 


0^89 





3-1 


3-48 


~ 






92 



APPLIED MECHANICS. 



Let the student show that = 0505 1 + 2-62, o= 0-52 
B + 3"1, B = 0-98 I — -89. Let him also fill in the columns 
o/i and o/b. Prove that 1 lb. of oil per hour means 8*27 hoi'se- 
power actually supplied, and fill in the column of efficiency. 

75. Accept the following measurements. A steam-engine 
employed in driving a dynamo macliine delivering electric 
energy to customers, each load being kept steady for four 
hours, each measurement being the average of the results 
obtained during the four hours, i is indicated horse-power ; 
B the brake horse-power measured by a transmission dynamo- 
meter ; the electrical horse-power e is obtained by multiplying 
amperes and volts to get the power in watts and dividing by 
746 ; c is the coal used per hour ; and w is the weight of steam 
used by the engine per hour. The governor acted upon the 
throttle-valve, and not upon the cut-ofK 



I 


B 


Amperes. 


Volts. 


E 


w 


c 


190 


163 


1,050 


100 


143 


4805 


730 


142 


115 


730 


100 


96 


3770 


544 


108 


86 


506 


100 


69 


3080 


387 


65 


43 


219 


100 


29 


21 00 


218 


19 





— 


— 


— 


1220 


— 



First plot the values of i and w, i and b, e and b, i and c 
on squared paper. It will be found that there is approxi- 
mately a linear law in every case. See if you get some such 
laws as — 

w = 800 -f 21 1 

B = 95 I - 18 

E = -QSb - 10 

c = 4-2 I - 62 



Now produce a few more columns of numbers and study them. 
Give w 4- I and c -=- i. Give w -j- b and c ^ b. Give w -j- e 
and c -f E. Also give w -^ c. 

76. Students may compare the above results with the 
following average measurements made at an electric supply 
station in 1891 : — 



APPLIED MECHANICS. 



93 





X 1 B 
1 


w 


c 


Average for 7 hours. 
11 a.m to 6 p.m. 


80-3 


571 


3,268 


552 


Average for 6 hours. 
6 p.m. to midnight. 


227-7 


163-2 


7,122 


742 


Average for 1 1 hours. 
Midnight to 11 a.m. 


37 


23-64 


2,143 


232 


, 24 hours. 
11 a.m. to 11 a.m. 


97-3 


68-3 


3,718 


453 



Here it will be found that although the load was constantly 
vaiying even when the averages for the 24 hours are taken with 
the others, we have linear laws between i, e and w, w = 1150 + 
26-25 I, and e =-72 i — 2. But c does not follow a linear law 
with the others. The reason lies in the fact that a spare boiler 
was used during part of the time, and there is consequently a 
greater consumption of fuel than if one or two boilers had been 
used the whole time. Since we have considered fuel consumption 
in the above exercises, it may not be out of place to introduce 
here some figures from the testing of a water- tube boiler : — 



steam per hour fioin 

and at 100° C. per lb. 

of coal. 


Coal per sq. ft. of 
grate per hour. 


Water evaporated per 
sq. ft. of total boiler 
heating surface per 
hoar. This is not re- 
duced to 100° C. 


V). 


13-40 
12-48 
12-00 
10-29 


7'74 
18-6 
29-8 
66-8 


1-24 

3-20 

' 4-70 

8-50 


103 
233 
357 

686 



li w = steam per hour per square foot of grate, / = fuel 
per hour per square foot of grate. Plotting w and /'on squared 
paper, we find a fair approach to a linear law, 

^ = 45 + 9-78/ 
45 



w 



/-/ 



+ 9-78. 



94 



APPLIED MECHANICS. 



77. If c is the total cost per hour when the useful horse-power 
p is being sent out by an hydraulic or electric or other supply com- 
pany (c includes interest and depreciation on first cost, rent, taxes, 
repairs, wages, stores, coal, water, otfice expenses and management), 
and if it is found that there is a simple law like c = av + b. where 
a and b are constants, prove that the average cost per hour is 
calculated from the average power in exactly the same way as the 
real cost c in any hour is from, the p during that hour. For if t is 
time in hoiu^s, then the cost during the year is, if t is the number 
of hours in a year. 



I c.dt = I (« P + h) (It = a\v 
Jo Jo *) 



vclt 




+ iT. 



Hence the average cost per hour is 



dt + h. 



Now 



'4r 



is the average power, call it p^, and we see that 



the average cost is a p,„ + b. The average power delivered in a 

day or year, di"sdded by the maximum jDower, is called the daily or 

yearly load factor. If / is the load factor and Pj is the maximum 

power, then average cost per hour = a p^ -^ / + i, 

78. An electrical company has arranged for a maximum output of 

1,000 horse-power. It is found that the total cost per hour in pence c is 

c = 08 p + 350. lip pence is charged for every horse-power hour sent 

out, what is the yearly profit when the average power sent out day and 

night is Tm 1 

Ans. — Subtract from p Fm the average cost j)er hour to the comjDany 
which is 0'8 p,^ + 350, and the profit per hour is (p - O'S) 7-^ - 350. 
The profit per annum in poimds is therefore this divided by 36-5. 

Exercise. — What charge per horse- j)ower horn- will give just no profit ? 

Ans. — X = 0-8 + SoO/Pm. 
Thus if Pm has the foUowine: values, we have the values of x 



Fm 


1,000 


500 


200 


100 


.V 


M5 


1-50 


2-55 


4-30 



79. We have seen that if a force e lb., acting throixgh e feet, 
overcomes a force r lb., acting through r feet, instead of the non- 

frictional law b = — .... (1), we find experimentally some such 

law as 

E = aR + A. 
It is evident that the constant a is a frictional resistance from the 
mere weight of the parts of the machine ; a is always found to be 



APPLIED MECHANICS. 95 

greater than -. The work done by e in the distance e feet ie 

E e or (?a R + A <?, and the useful work is r r, so that the 

r 

„„ . rR ea 

Jliinciency x = 



^« R + A e 1 I •*• 
aR 

It will be observed that the lare-er r is, the smaller is — . or 

av: 

the more insignificant is the term due to weight of parts of 

machine, and the more nearly does the denominator approach 

unity. However gxeat r may become, the efiiciency cannot 

Y \ . V 

exceed . - ; and as a is always greater than -, the efficiency is 

always less than unity, as was to be expected. 

1 r 
Denote a by v • ■ , where k is always less than 1, Then the 

efficiency can never be greater than k. If e and r are 

feet per second, then ^^ and ^^ represent the horse-powers. Let 

us call ^^ the total borse-power t, and ^^ the useful horse-power 

u, and we shall call t — u the lost or wasted horse-power l. Then as 
El? ae 'R.r Ae 1 , Ae , , Ak e 

80. It is interesting to know, from such examples as we have 
studied in Arts. 73-76, that if t is the indicated horse-power of 
a steam-engine, and u is the brake horse-power, or what is 
given out usefully by the crank shaft, and if this shaft drives a 
dynamo machine, and e is the electrical horse-power given out ; or 
if the shaft drives a pump, and e is the effective horse-power of 
the pump, we always find (probably with only approximate 
accuracy) a linear law connecting t and u, and tj and e, and 
therefore also connecting t and e. Furthermore, in a steam- 
engine which does not vary its period of cut-off — that is, the 
regulation is by throttling or the boiler-pressure changes — if w is 
the poundage of steam per bour, there is a linear law connecting 
any two of the quantities w, t, u, and e. 

If we write (1) as 

u = A;t— a, orT= rv -\- =-a, 

it is curious that in so many of our machines in which the 
transmission of power is by mere shafting k should be so much less 
than unity as it is. 

This occurs because a me.-'-e torque is very seldom applied to a 
shaft when power is being given or taken. When the power is 
supplied or taken off by a belt, the power is proportional to the 
difference of pulls in the belt, whereas the loads on the bearings 
and the friction are nearly constant. Such kinds of driving tend 
rather to increase a than to diminish k. All spur or bevel gearings 



96 APPLIED MECHANICS. 

tend to diminish. X-, and only by their mere weight do they tend to 
increase a. The poor efficiency of ordinary lines of shafting is a 
matter to which few people seem to have paid any attention ; and 
this is, I think, altogether due to the ahsence of dynamometer 
couplings or other means of forcing upon the attention the actual 
amounts of power which are heing transmitted at each place. 
Electrical engineers have very accurate methods of measuring the 
power giren out by their dynamos, and hence they are concerned 
as to the actual mechanical power supplied to them. It was 
almost altogether due to this that direct driying of dynamos from 
engines took the place of indirect methods ; and, indeed, I may 
go further, and say that it was due to this that the great improve- 
ments have taken plaoe in steam-engine manufacture and working 
during the last fifteen years. If even the practical engineer is 
beg-inning, therefore, to think of economy, it is greatly due to the 
fact that electricity is paid for at so much per unit of energy; 
although, no doubt, much is also due to the fact that economy of 
coal is very important in ocean-going steamships. 

81. In any machine in which a small effort, e, overcomes a great 
resistance, r, we usually fimd that the frictional loss of energy in 
the machine is almost altogether dependent upon r. Thus in the 
inclined plane (Fig. 25), with e parallel to the plane, the friction 
is independent of e. T\Tien in the inclined plane of small inclina- 
tion E is horizontal, the friction is almost altogether dependent 
on the load to be lifted ; and it is so in a screw-jack and in the 
differential pulley -block. It is not so much the case in cranes, 
unless such gearing as wonn-gearing is employed. 

Let us, then, suppose that the loss of energy in the machine is 
altogether due to r. Then in the direct use of the machine — that 
is, E overcoming r— if the efficiency is only 50 per cent., the loss 
of energy is equal to r r (if r is Hfted r feet), r r being also the 
useful energy . Consequently, if we try by increasing r to R^ to 
make the machine reverse, the loss is always R^r; or the whole 
energy would be wasted if the machine could be supposed to move. 
If, then, the direct efficiency is equal to or less than 50 per cent., 
the machine cannot reverse or overhaul. 

In any machine there is some loss due to e, and also to the 
weight of the parts of the machine, as well as to r, and conse- 
quently the direct efficiency must always be less than 50 per cent, 
if the machine is not to overhaul. On the assumption that the 
lost energy, l, is 

r, = mvi r -\- n 'E e -\- A, 

where a is a constant, due to the weight of parts of the machine, 
and in and n are fi-actions, we have, in direct working, 

K e — R r = L, 

Ee — Rr = 7;iR?--|-«Ee-|-A, 

^^^(7w4jOjE^-fA 
1 — « 

Hence the dii^ect efficiency v is -. — \ ~ ' — . — .... (1), 
^ (w + 1) R ;• -I- A ^ " 



APPLIED MECHANICS. 97 

)v — : -; — . Asain, in reversed workinff 

*n-\-l + a/r r ° ' ° 

Kir — E (9 = L, 

B}r — -Ee = m R^r + « e <? + A, 
B}r (I —m) — A 

The reversed efficiency y = — r- = ^ — 77 — —. — t-— .... (2). 

•^ ^ R^r {I + n) rV ^ ' 

This is or negative when 

~^- + m = or > 1. 

R^r 

As r1 may he made very great, our condition of non-reversihility 

must he «i = 1 or > 1, Hence the direct efficiency must he equal 

1 — n 
to or less than r— — ; — . We therefore see that there is no general 

2 + a/r r ^ 

rule such as many people seem to helieve in — the result of mis- 
leading mathematics. We can hardly call it a general rule to say 
that, if VI is equal to 1, r cannot overcome e, because m==l means 
that all r's energy would bo wasted. 



98 



CHAPTER VI. 



MACHINES. SPECIAL CASES. 



82. Blocks and Tackle. — It is very good to have a general 
law telling us about machines in which there is no friction. 
That law you now know. 7'he mechan- 
ical icork given to a machine is equal to 
the v:ork given out by it, unless it is 
stored up in the machine itself by tlie 
coiling of a spring or in some other way. 
But, besides knowing the law itself, it is 
well to know what it leads to in certain 
special cases. Take, for instance, a 
pulley-blcck, Fig. 38. It is evident 
here that if we have three pulleys in 
the block b, if the effort p acts through 
six inches, w will only rise one inch, 
and therefore P will balance six times 
its weight at w if there is no friction. 
The mechanical advantage is therefore 
six. 

This is a case in wMch. it is less dif- 
ficult tlian usual to trace the loss of 
energy due to friction. If, when the pull 
of a cord is p^ on the overcoming side of a 
pulley, the tension on the other side is 
given by Pi = a p^, — b, where a is less 
than 1, and b does not depend much on the 
weight of a sheave, but rather on the vis- 
cosity of the roi^e. we may take it that 
Pg =^ « Pj — b, and so on. I shall leave it to students to work 
out an algebraic expression connecting w, the sum of Pj, Pg, P3, 
etc. , and Pq. w will include the weight of the lower block. But 
it is good to take a nimierical example. For instance, let 
Pj = 0*9 P(j - 3, the forces being in pounds. Then 




Fig. 88. 



P2 = 0-81 
Pa = 0-729 



2-7 - 3 
0-13 - 3 



•6561 p„ 



Pg = -531 5 Po - 
^nce w = 4"2171 Pg 



7-317 

9-288 

11-06 



0-81 Po 
0-729 p, 
■6561 p„ 



o-/, 
8-13, 
10-32, 
•5905 Pn — 12-29, 



3 = 
3 = 
3 = -5315 p" 



14-06; 
53-50, or P. = 0^237 w + 12-7. 



APPLIED MECHANICS. 



99 



llie efficiency x = 



•167 w 



•705 



= 53^6' 



237 w + 12-7 ~ 1 -{- 



Hence, however great w may te, the efficiency is less than 70*5 
per cent. When w = 53*6 the efficiency is only half this. 

83. This is a fairly good example of the cumulative effect 
Df friction. If we sfive 



Po 



wer Pn to a machine 




whose efficiency is Cq, 
if this machine gives 
power to another whose 
efficiency is e-^ and so 
on, the power given 
out by the last of a 
series of machines is 

^ C\ /N ^A ^ ^1 ^ Cfrt X t> CO. 

If, as in transmitting 

power for a great dis- p. gg 

tance by means of ropes, 

all the contrivances are the same, we have the compound 

interest law of diminution of power. (See Exercise, page 232.) 

Students may find it interesting to study a machine or method 

of transmission of power in a manner allied to what follows 

in Art. 91 of this chapter. 

84. Inclined Plane. — Again, take the inclined plane, 
Fig. 39 ; w is a weight Avhich may roll down the plane without 
friction, let us suppose; p is the pull in a cord which just 
prevents w from falling. The cord is parallel to the plane. 
Evidently when w rises from level a to level b the cord is 
pulled the distance A b ', that is, w multiplied by the height of 
the plane is equal to P multiplied by the length of the plane. 
Thus, if w is 1,000 lbs., and the length of the plane 10 feet for 
a rise of 2 feet, then ten times p is equal to 2,000, or p is 
200 lbs. 

Barrels and boilers are often raised along an inclined plane, 
ropes or chains held fast at the top of the plane passing round 
the cylindric object and back towards the top of the plane 
where force is applied to them. In this case p is evidently 
only half what is stated above. As the weight rolls on tlie 
double rope or chain there is not much friction. In Art. 90 
we shall consider the effect of friction in the inclined plane, 

85. The Sc^ew. — Again, suppose there is no friction in the 



100 



APPLIED MECHANICS. 




Fig. 40. 



screw A B, Fig. iO ; if rfc rises it lifts a weight say of 3,000 lbs. 
Now, if the screw make one turn it rises by a distance equal 
to its pitch; that is, the distance (measured parallel to the axis) 

between two threads. 
Say that the i)itch 
is "02 foot, then 
when the screw 
makes one tura it 
does work on the 
weight 3,000 x -02, 
or 60 foot-pounds. 
But to do this, p 
must fall through a 
distance equal to 
the circumference 
of the pulley a, 
about which I sup- 
pose the cord to be 
wound. Suppose 
the circumference 
of the pulley to be 
6 feet, then p multiplied by 6 must be 60, or p is 10 lbs. The 
rule, then, for a screw is this — effort multiplied hy circum- 
ference of the pulley equals resistance multiplied hy pitch of 
screiv. It is not usual to have a pulley and a cord working 
a screw; it is more usual to have a handle, and to push or 
pull at right angles to the handle. Instead of the circum- 
ference of the pulley, we should take, then, the circumference 
of the circle described by the point where the effort is applied 
to the handle. 

Example. — A steam-engine gives to a propeller shaft in 
one revolution 60,000 foot-pounds of work ; the pitch of the 
screw is 12 feet. What is the resistance to the motion of the 
vessel? Answer: The resistance in pounds multiplied by 12 
gives the work done in overcoming this resistance^ and this 
work must be equal to 60,000 foot-pounds ; hence the resistance 
to the motion of the A^essel is 5,000 lbs. [It would be more 
correct to say that this is the work done per foot-travel of the 
vessel, assuming no slip of the screw.] 

Screws are used for many purposes. When used, as in bolts 
to fasten things, the threads are triangular in section. The 
Whitworth thread is shown in Fig. 41. a b is the pitch ; G H 



APPLIED MECHANICS. 



101 



is -96 of the pitch, the angle h b j being 55° ; the corners are 
rounded, so that i E is '64 of the pitch. The Sellers thread 
used in America has an angle of 60°. JFig. 42 shows a square 
thread. There is less friction and less wear with this form of 
screw, and it is used when accuracy of motion is important. 
As there is only half the amount of material resisting 
shearing, the square thread has only half the strength 

of a triangular thread. The Buttress thread 

of Fig. 43 has strength and accuracy as to 

motion; it is used when 

the important motion is ^ y,,,,/' 

in one direction only. t'^'" 1 V 

A student has plenty <cu..^y,, i r% 

of opportunity of examin- o " i^^'- 

ing examples of the use of "p^ X^^ 

screws. To fasten things 

together we have many 

kinds of bolts and many 

forms of heads and nuts, 

and many ways of locking 

nuts. Other fastenings 

also, such as pins and 

keys and cottars, come 

before his eyes every day Fig. 42. Fig. 43. 

in the workshops, and he 
must become familiar with them and their usual proportions 
both in the shops and the drawing-office. He must make 
a very careful examination of such a tool as a good screw- 
cutting lathe, and especially of the mechanism of the slide 
rest. The calculation of the proper change wheels for the 
cutting of a particular screw is a very simple matter ; but all 
so simple as it is, he must work out some examples. 

He must note the shapes of threads of screws for wood, for 
sand or mud, and for water ; the use of screw piles as the 
supports of structures, the forms of screw-propeller for steam- 
ships and wind -mills, and screws used as fans for blowing air. 

86. Wheel and Axle. — If a and b, Fig. 44, are two pulleys or 
drums on the same axis and having cords round them, a small 
weight, p, hung from a, will balance a larger weight, w, hung 
fi'om B. For, suppose that one complete turn is given to the 
axis, P falls a distance equal to the circumference of A whilst 
w is rising a distance equal to the circumference of b. Hence 




Fig 41. 



103 



APPLIED MECHANICS. 



p X circumference of A = w x circumference of B, 
or, what really comes to the same thing, 

p X diameter of a = w x diameter of b . . . . (1), 

or p X radius of A = w x radius of b. 

In practice A is often a handle and B a sort of barrel on 

v/hich a chain may be wound. Or the axle may be compound, 

consisting of two parts, the diameters d and 

^^^ d being nearly equal, the rope being coiled 

^Hi||||j\ B round them in opposite dii-ections so as to 

^^H^y^te form a loop, upon which hangs a pulley. In 

^^■I^Bp this case (1) becomes 

^''^^ P X diameter of a = w x c)~ 

Or, again, a may not be on the same shaft 
as the barrel b, but may gear with it through 
spur and bevel-gearing. Thus, in a hand 
crane, the handle may turn many times for 
one turn of the barrel. Also there may be 
a snatch-block, so that when two feet of chain 
are coiled on the barrel, the weight rises 
only one foot. There is no mystery about 
tooth gearing, and anyone who has looked at 
a crane knows how by merely measuring the 
length of the handle, or, rather, the circum- 
ference of the circle described by the handle, 
and the amount of chain coiled on in one 
reyolution of the barrel (this is larger than the 
circumference of the barrel itself), and counting the nimiber of 
teeth of the dri\"ing and driyen wheels, what is the velocity 
ratio, and, therefore, the hypothetical mechanical advantage. 
I write for men who go about among machinery, and the most 
illiterate workman knows well how speeds of shafts depend 
upon numbers of teeth. Also, all my readers haye seen com- 
binations of ban-els, snatch-blocks, blocks and tackle and crab- 
winches. A worm and worm wheel, or other kinds of screw 
gearing may also be imagined. 

Example. — The handle of a crab or crane is 18 inches long; 
20 inches of chain are wound on in one reyolution of the 
barrel. The barrel is driyen from the handle by a train of 
wheels — driver 15 teeth, follower 64 teeth; driyer 16 teeth, 
follower 60 teeth (the value of this train is said to be 




Fig. M 



APPLIED MECHANICS. 



103 



15 X 16 ] 

wj 7.-A, orvp ); so that the handle makes 16 revolutions for 

one of the barrel ; the chain lifts the weight through the 
agency of a block and tackle, with 3 sheaves below. 

When the barrel turns once, 20 inches of chain are coiled 

20 
on, and therefore the weight rises -^ — ^ inches or 3J inches. 

The handle turns 16 times, and the hand 

moves through 16 x 36 tt inches. Hence 

the velocity ratio of hand to weight is 

16x367r ^^^ ^, 

Kj = 543. Orten we have a means 

provided of disengaging the spur wheels, 
driver 16, follower 60, and so we can de- 
crease the velocity ratio to 543 x -— or 
•^ 60 

about 145. 

87. A differential pulley-block is shown 
in Fig. 45. When the chain e is pulled, 
it turns the two pulleys, or rather one 
pulley with two grooves, b and c. Now c 
is a little smaller than b, so that, although 
at D the chain is lifted, it is lowered at f. 
If the circumference of b is 2 feet and that 
of c is 1*95 feet, then, v/hen e is pulled 2 
feet, D is lifted 2 feet, but f is lowered 
1*95 feet, so that there is 0-05 foot of chain 
less than before in the parts d and f, and 
the pulley g rises the half of this, or "025 
foot. If R is 2,000 lbs., then 2,000 x -025, 
or 50, must be equal to the pull e multiplied 
by 2, hence e is 25 lbs., or an effort of 
25 lbs. is able to overcome a resistance of 
2,000 lbs. The general rule, then, for the 
differential pulley-block is, effort e multi- 
plied by circumference of larger groove b is 
equal to resistance r multiplied by half the 
difference between the circumference of the two 
grooves b and c. You will find that this rule comes to the 
same thing — effort multiijlied by diameter of B is equal to 
resistance multiplied by half the difference betiueen the diameters 



104 



APPLIED MECHANICS. 



of B and c. The grooves are furnished with ridges to catch 
the links of the chain, so that there shall be no slipping. 

It is only when we experimentally measure the effort E 
which will slowly overcome the resistance R in this piilley-block 
as actually made, that we see how great is the f rictional waste of 
energy. It is in consequence of this that, however great the re- 
sistance R may be, it ^ill not fall, even when there is no force 
exerted at e. This property of not "overhauling" makes the 
differential pulley- block the very useful implement which we 
know it to be in' a machine shop. It is the characteristic of 
any machine which has a very great velocity ratio that if its 




Fig. 46 



Fig 48. 



efficiency is less than half, it will not overhaul (see Art. 81), 
and lifting machines which do not overhaul are often very 
convenient. 

88. Equilibrium in one Position.— In all the machines 
which we have hitherto considered, vv^e could give motion without 
altering the balance of the forces, but there are many raachines 
in which the mechanical advantage alters svhen motion is given. 
In such cases you will employ your general principle, but you 
must make your calculation from a very small motion indeed. 
For instance, in the inclined plane, if the cord which prevents 
the weight from falling is not parallel to the plane— say that 
it is like m, Fig. 46— you will find that the necessary j)ull 
depends on the angle the cord makes with the plane. ^ Now, 
suppose that the cord pulls the carriage from h to c, evidently 
the angle of the cord alters. The question is, what is P, that 
it may support w in the position shown in the figure ? We 



APPLIED MECaANiCS. 105 

know tliat it will be different after a little motion, but what is 
it now ? Imagine such a very small motion from 6 to c to 
occur that the angle of the cord does not alter perceptibly, and 
now make a magnified drawing, Fig, 47. p has not fallen so 
much as the distance h c, it has only fallen the distance h a {c a 
is perpendicular to h a). In the meantime the weight w has 
been lifted the distance k c. Hence, 

wxk c ought to be equal to Pxb a. 

Thus^ if you measure k c and b a on your magnified drawing to 
any scale you will find the relation between p and w. Another 
way of finding the same relationship is this. We know that 
the weight of w acting downwards, the pull in the cord, and a 
force acting at right angles to the plane, are the three forces 
which keep w where it is. Draw a triangle whose three sides 
are parallel to the directions of these three forces, Fig. 48, 
with arrow-heads circuital ; then x and y are in the propor- 
tion of w and p. Here we have used the principle called 
" the triangle of f jrces " to find p. 

EXERCISES. 

1. Find the force parallel to the plane required to draw a weight of 
2 cwt. up a smooth inclined plane. Height of plane, 3 ; length, 5. 

Ans., 1-2 cwts. 

2. In a screw-jack the pitch of the screw is | inch ; radius of circle 
described by hand', 19 inches ; find the velocity ratio. It is found that a 
force A at tlie handle of 30 lbs. will overcome a weight of 2,300 lbs., and 
one of 10 lbs. will overcome a weight of 500 lbs. ; what law connects A 
and w ? When \v is 3,000 lbs., what is a? What is the efficiency ? 

Ans., 318-47 ; a = ^ + 4^ ; 37^ lbs. ; 25 per cent. 

3. The handle of a lifting-jack measures 24 inches in length; the 
pitch of the screw is f inch ; what force applied at the end of the handle 
would be required to raise a load of 22 cwt., the effect of friction being- 
neglected? Ans., 6-125 lbs. 

4. Pitch of screw-propeller, 18 feet; slip, 10 per cent. ; speed of shij), 
15 knots; find the revolutions per minute. What is the thi-ust if the 
actual horse-power spent bv the propeller is 2,000, and the waste by surface 
friction is 30 per cent ? What is the torque in the shaft ? 

Ans., 93-83, 13-57 tons ; 50 ton-feet. 

5. The British Association rules for the pitch and diameter and index 
number of screw-threads for instrument work are 

d = 6 pi, p = {0-9f- 
Taking the values of 0, 1, 2, 3, .... 12 for s, calculate p and d, and keep 
for reference in a table. Try to what extent the rule p = -08 d -\- -Oi for 
the pitch and outside diameter of triangidar screws agrees with the well- 
known Whit worth table. 



lOG APPLIED MECHANICS. 

6. What must te the difference in the diameters of a comijomid wheei 
and axle so that the velocity of e may be eighty times that of r, the 
length of the handle being 2 "feet ? Ans., 12 inches. 

7. The ^veight on a crane is carried by a snatch-block ; the chain goes 
to a barrel on which 17 inches are wrapped in one revolution. The 
barrel is driven by wheels, which give it 1 revolution for 15 revolutions 
of the handle, and the hand describes a circle of 21 inches radius ; what is 
the velocity i^tio ? Ans., 232-8. 

8. In a single-purchase crab the pinion has 12 teeth and the wheel 
has 78 teeth, the diameter of the barrel (or rather of the chain on the 
barrel) being 7 inches, and the length of the lever-handle 14 inches. It 
is found that the application of a force of 15 lbs. at the end of the handle 
suffices to raise a weight of 280 lbs. ; find the efficiency of the machine. 

Ans., 0-718. 

9. In the differential pulley, if the weight is to be raised at the rate 
of 5 feet per minute, and the diameters of the pulleys of the compound 
sheave are 7 and 8 inches, at what rate must the chain be hauled ? 

Ans., 80 feet per minute. 

10. In a differential pulley in which the velocity ratio of fall to lift is 
30, a pull of 7 lbs. will just raise a load of 24 lbs., and a pull of 25 lbs. a 
load of 240 lbs. Find the pull required to lift 600 lbs., and the efficiency 
of the machine when such a weight is being raised. 

Ans., 55 lbs. ; 0-36. 
89. "When one body touches another, and there is equilibrium, if 
there is no friction, this means that there is no tangential force at 
the surface. The force with which either body acts on the other 
has no tangential component. If, however, there is friction, and 
we assimie that the fiiction is ^r where r is the force normal to 
the rubbing surface, it is evident that the total force at the place 
makes an angle <p with the normal if tan. <}> = fi. If the total 
force makes an angle less than (p, its tangential component is less 
than the friction, and there can be no 
motion. If, then, two bodies a o b and 
DOC touch at o, and >' o x^ is their com- 
mon normal at o, so long as the direction 
of the force acting between the bodies lies 
inside the cone a o p, whose axis is o ?r, 
the angle p o n or q o n being (p, there will 
be no sliding motion or rubbing at o. 
Thus, when the block lies en the inclined 
plane (Fig. 50), the total force ti-ans- 
mitted between block and plane is w, the 
vertical weight of the block. So long 
as the angle is less than <p, there is no 
sliding. But the angle is the angle of 
the pkne ; hence sliding is about to begin 
when we are increasing the angle of the plane, and it has become 
as much as (p. 

If the body d o c is at rest, and A o b mores, touching the first, 
the motion is one of sliding, rolling, or spinning, or combinations 
of these three. If there is no sliding, so that o is momentarily at 
rest, the instantaneous motion of a o b must be an angular velocity 




APPLIED MECHANICS. 



107 



7 about some axis through o. If this axis makes an angle with 
the tangent plane, there is an angular velocity of rolling y cos. 




Fig. 50. 



about an axis in the tangent plane, and an angular velocity of 
spinning 7 sin. 9 about the common normal to the two planes. 

If in Fig. 49 the surfaces are cylindric and the rolling angular 
velocity is o), and if r is the radius of cm-vature of a o b at o, and 
if R is the radius of curvature at o of the fixed surface doc, and if 
V is the linear velocity of the point of contact, it is easy to show- 
that 



\r R/ 



The centres of curvature are supposed to he on different sides of 
the tangent plane. 

If in Fig. 50 there is fi'iction between w and the plane, and 
motion up the plane is steadily taking place, draw k, making an 
angle n o r = (/>, with o n the normal. Let p 
show the direction of the pulling force, and w 
of the weight. Knowing only w to calculate p 
and R, we have simply to use the triangle of 
forces. We are given one side and the angles 
to find the other sides. Thus draw ab repre- 
senting w to scale ; and di-aw b c an i a c 
parallel to the two unknown forces, with the 
arrow-heads going circuitally. Measure b g 
and c A, and the forces p and r are known. 

Analytically. — Resolve all the forces horizo7i- 
tally and vertically, and we have (see Art. 31), if 
the angle pod is a, as p makes an angle a + 6 
with the horizontal, and r makes an angle r o h, which is the 
complement of [cp + 0), 

p COS. (a + 0) = R sin. (0 + 0) .... (1) 

p sin. (a + 0) -h R COS. <^ + 0) = w . . . . (2). 

Ab we really do not wish to know r, use its value from (1) in (2), 




108 APPLIED MECHANICS. 

and we find p sin. 'a + 0) + cos. > + 0) L£!i!?i_f5 J = ^^ 

sin. {(p -\- a) 
w sin. {<!> + 6) 



COS. (</) + d) COS. (a + 0) + sin. ((|) 4- 9) sin. (a -f 0) 
w sin. {<p + 0) _^ ^ sin. (0 + cb) 
COS. J(a+0)-(4, + e)[ cos.(a-<^) 



If p is the force required to Just prevent the body sliding doicn 
the plane, it is necessary to draw the angle nor upwards, instead 
of downwards, and to take — ^ for <p in the formula. Then 

COS. {a+ (p) ^ ^ 

Example 1. — If a ^ o, p = w — ^-^^ for upward motion ; 

COS. <p 

p = w — — for downward motion, p is for downward 

COS. (p 

motion when e = (p. Under these circiunstances the body wiQ 
just be about sliding down under the action of its own weight. 

When e = <p, it takes a force p = 2 w sin. (p to drag the body 
Tip the plane. 

Notice that when the body is pulled up the plane, if is a 
small angle, expanding sin. (<^ + 0), and di^iding, we find p = w 
(tan. (p cos. + sin. 0). 

Taking cos. = 1, we have, sinc« tan. <p = fi, p = ^"W-i-w 
sin. 0. 

Now juw is the force that must be exerted if we have no 
inclination 0, and w sin. is the force that must be exerted if we 
have inclination but no friction. Hence the rule usually employed 
in calculating the pulling force on a vehicle (Art. 47) — namely, 

the total pull when the inclination is 1 in w (or sin. = - j — is 

equal to the pull necessary on a level road plus - th of the weight. 

It is only true when is small. 

Example 2. — If a = — 0, so that p acts horizontally, p ^ w 

Bin. {(p + e) ^ /^ , - n . ^. sin. (0 - cp) 

p- rf = w tan. (0 + (p) for upward motion, p = w j-z — 

cos. {(p + e) \ i ^j 1 5 (,Qg_ ^Q _ ^ 

= w. tan. (0 — <^) for dowmvard motion. 

Example 3. — If a = o, so that there is no friction, p = w ^- — '- — . 
^ ^ ' ' cos. a 

Example 4. — Find a so that p for upward motion may be a 

minimum. That is, what value of a will cause — ^-7 — to be 

cos. (a — <p>) 

a minunum ? That is, wliat value of a will make cos. (a - <p) 

a maximum ? EWdently it is a maximum when a — <p = ^ 

or a = (/), and then p = w sin. (<^ + 0). This important result 

seems to be completely ignored by men who deal practically with 

traction j)roblems. 

Example 0. — If there is no force p, what is the acceleration 



APPLIED MECHANICS. 109 

down the plane ? We must first find what vahie of p acting 
parallel to the plane would just be overcome. This is given in 

Example 1 as w — — —. Now, this force acts upon the mass 

^ cos. (/) ^ 

w 

— , and acceleration is force -^ mass ; so that the acceleration is 

q — Li zl^ Of course, when A = 0, this becomes the well- 

^ cos. 5 f 5 

known g sin. Q. 

Exam2)le 6. — ^ATien the nut of a square-threaded screw is 
turned, the surface of the thread is like an inclined plane, 

the tangent of whose inclination, Q, is - — j , if <? is the mean 

diameter of the thread (or the diameter of the pitch cylinder of the 
screw). The nut presses upon the inclined plane exactly as in 
Example 2, and hence p = w tan. (0 + 4)) .... (1) if p over- 
comes w, the total v/eight to be lifted when the nut is turned ; or 
p = w tan. (0 — (/))... . (2) if w overcomes p. If the length 
of the handle to which the real force a is applied be I, then 

p d 
lA = ^vd or A = ^y; so that (I) and (2) become 

A = i|wtan. (0 + </>)....(!), 

A = I ^ w tan. (0 - </>) (2). 

(1) and (2) are equal, of course, if ^ is 0, and then 

A = ^ -, w tan. d. 

In this case, as there is no friction, the efficiency is 1, and the 

mechanical advantage is — = (3). The mechanical 

A d tan, Q 

2 I 

advantage from (1) when there is friction is -tt ttt-, — ^ .... (3). 

° ^ ^ d tan. (e -\-(p) ^ ^ 

Hence the efficiency when there is friction is e =: , — '—-^ .... f4). 

tan.(0 + (|)) ^ ' 

When w overcomes A the efficiency is similarly e= ^^^ ~^^ . . (5). 

Both (4) and (5) become 1 when there is no friction. 

It is an easy exercise in the calculus to show that (4) is a 

maximum when = _ — -^ ^ . . . . (6), and that (5) is a maximum 

when = ^ + i . . . . (7). 

If, then, a screw is to work as much in one way as the other, it 
seems reasonable to use = — or 45° as the angle of its mean 

spiral. 

Note that when Q is as small as <p, w cannot overcome a, and 
the screw will not overhaul or reverse. In this case, if a over- 



110 APPLItD MECHANIC-S. 

tan. o TT-- 

comes vr,e= — - . \Vlie::i a lubricant is used we cannot be 

tan. 1 G> 

sure that tlie here assuiiir'i b.w .:f fricrlrn holds ; that is, /x is not 
a constant. When a lu":r::ai:t is nit use-i it is saie to say that in 
no actual screw-jack does '.ne find t? tipiM aching * in ^-alue. 

!N ote that when there is no friction the mechanical advantage 
• ^ 2/ - . « pitch ^. ^ 

IS — = -T-r 1, where tan. 6 = - — — : so that 

A rftan. e' rd 

W ^ -lird ^ 2t/ 
A rf X pitch pitch' 
or cireumfereHee described hy the end of the handk 4- pitch, the 
rule giren already. 

I have not considered the very DZ-nsiderable Ic^s of efficiency 

due to friction because the load which is lifted is being kept from 

turning. Eren the carefully-constmcted and well-lubricated square* 

threadei screw-jack in my laboratory has an ef5.ciency of only 

about ■25 eren at the highest loads. The ordinary jack used in 

workshops has a xery much smaller efficiency than this. When 

the screw has a triangular thread we may assume that with the 

same kinds of rubbing surface the 

Q co-efficient of friction has greatly 

^ increased ; in reality it is the normail 

^ ^^^-^^ ^ pressnre on the thread which h;\s 

J^ incTt^sed. 

^r-M^^ -- - 90. "When ai>i/ffiw^ force f, making 

^ ^■^^ ^ an angle /8 with, the normal, is applied 
: to more a block which is being pressed 

against a surface, the tangential 
component which overcomes friction 
F:g. 52. is F sin. /3. The normal component 

p cos. ^ diminishes r, so that the fric- 
tion is ,u (e — F COS. 3). Hence, when sliding occurs, 
F sin. ^ := u (k — f cos. /8), 

^" ^=^ sin.3-t-Me>^-/ 8 •••: ^^' 
For any given value of /t it is easy to find the Talue of /3 which 
will make p a miTii-minn- In fact, the denominator of (1) is a 
Twa-dTmiTn when COS. /8 = u sin. ;3, or fi tan. ^ = 1, or ^ is the 
complement of *, the angle of repose. 

But if F is a pushing force, we have f sin. /3 = fi (n -|- f cos. /8) ; 

so that r = -. v-^- -• In this case f becomes infinity 

;__ ,3 — M cos. /B ^ 

ii o is 1tS= :ii r. the complement of ^. For an angle /3 which 

is less rhar! 90" the pmhing s must be greater than the puUiing f. 

When one piece of machinery drives another at a sliding contact 

this great distinction between pushing and pulling must be 

remembered. 

EXEECISES. 

1. A weight of 5 cwt. resting on a horizontal plane requires a 

horizontal force of 100 lbs. to move it against friction. What, in that 

case, is the value of the co-effident of friction ? Atis., 018. 



APPLIED MECIIANJCS. 



Ill 



2. A weight of 50 lbs. is supported by friction alone on an inclined 

plane ; what is the force of friction ? Angle of plane, sin f . yins., 20 lbs. 

3. A body placed on a horizontal plane requires a horizontal force equal 
to one-half its own weight to overcome the friction. If the plane be gradually 
tilted, at what angle will the body begin to slide? Ans., 26° 34'. 

4. Find the force parallel to the plane required to draw a weight 
of 40 Ibs.fup a rough inclined plane ii fi = }, the inclination of the plane 
being such that a force of 12 lbs. acting at an angle of 15° to the plane 
would support the weight if the plane were smooth. A)is., 24*35 lbs. 

5. On a rough inclined plane it is found that a body is just supported 
on it by a horizontal force equal to three-quarters the weight of the body. 

Find the co-efficient of friction. Angle of plane, sin f . Ans., ^. 

6. Two unequal weights, Wj and W2, of the same material on a rough 
inclined plane are connected by a string which passes over a fixed pulley 
in plane. Find inclination of plane when the system is in equilibriirm. 

Ans., tan ~' ^ (^^^ + ^\ 

Wi - W2 

7. Two rough bodies, Wj and Wg, rest on an inclined plane and are 
connected by a string, the inclination of string to the horizontal being the 
same as the plane. If the co-efficients of friction are wi and /^g respectively, 
find the greatest inclination of the plane when the system is in 
equilibrium. ^^^^_^ ^^^ -1 A^iW^J^^.^ 

^^\ + Wa 

8. A plane is inclined to the hiorizontal at 24° ; there is a load of 
1,200 lbs. on it, the co-efficient of friction being -18. Determiae (1) the 
force required to just draw the load up the plane ; (2) the force required 
to just prevent it slipping down ; (3) the force required to support it if 
there be no friction, the direction of the line of action of the force being 
in each case 23° with the plane, and above it. Determine the correspond- 
ing results if the direction of the force be parallel to the plane, and again 
if the force be horizontal. Obtain also the least force, in magnitude and 
direction, which is necessary in each of the three cases fii'st referred to ; 
that is, for raising, lowering, and supporting without friction. There are 
twelve results in all. 

Ans., 





Effort E in pounds. 




E making 2.3" 
with plane. 


K parallel to 
plane. 


E 

horizontal. 


E a mini- 
mum. 


Upward motion. 


692 


685 


816 


674 


Downward motion. 


342 


291 


295 


286 


No friction. 


530 


488 


534 


488 



112 



APPLTED MECHANICS. 



9. A load of 1.200 Its. has to be pulled along a horizontal piano 
Determine in magnitude and direction the least force necessary to do this, 
the co-efl&cient of friction heing taken 0-4. Ans., 446 ; 2F 48'. 

10. Determine the efficiency in the case of a scre"sv 2^ inches diameter, 
in -v^'hich there are four threads to the inch, taking the co- efficient of 
friction -04. Ans., 0442. 

11. Taking the mean diameter of the threads of a 1 inch holt to he 
■92 inch, nimiber of thi-eads per inch 8, the co-efficient of friction reduced 
to an equivalent square thread 'IT, find the turning couple required to 
overcome an axial force of 2^ tons. Ans., -20 ; 46'1 pound-feet. 



JOINTS WITH FEICTIOX. 

91. When we know the resultant of the loads or forces on a pin 
or journal, and the law of friction, we can calculate the resultant 
force on the eye or step. Thus, if no re- 
presents the resultant load w upon the journal 
s p, and there is also a twisting couple turning 
the journal in the direction of the arrow t, 
the journal rides up in the step till p is the 
point of contact, and until o p q = <p is the 
angle of repose, p q heing a vcxtical force equal 
to w, and this is the resultant of the forces 
with which the step acts on the journal; /j., the 
coefficient of friction, is tan. o p q. 

It is usually stated in books that the resul- 
tant force between an eye and pin must act 
through some point such as p in the siu-face of contact. To show 
that thLs is a mistake, imagine a b and b c two pieces with a pin 
joint at B. These letters indicate the centres of pins at a, p^ and 




Fjg. 53. 




0. If we neglect the weights of the pieces, imagine frictionless pina 
at A and c, then any force communicated from a to c through the 
frictional joint b is in the straight line a c ; 
for the resultant force of the pin a upon the 
piece A B acts thi-ough the centre a, and the 
resultant force of the piece b c on the pin c 
acts through the centre c. Imagine the pin 
at B to be rigidly fixed to b c. The force 
between a b and c b at the joint b must 
be in the direction a c ; and we can have 
so much friction at b that a c does not 
Fig, 56. necessarily act through any point of tho 




APPLIED MECHANICS. 



113 



Biirf ice of contact. We cannot, however, imagine this happening if 
the pin and eye at b meet only at one point or line at right angles to 
the paper ; and it is therefore important that pins and eyes should be 
^f slack fit, and we shall in future imagine them to he so. V/e avoid 
unnecessary complication if we imagine that when two pieces act 
through a pin, the pin is rigidly attached to one of them. Let cd 





Fig. 56. 

and E F be two frictional pins acting on the piece g h. Let a and 
I? be their centres. Let the coefficients of friction be a and b, or 
the angles of repose cp and cp^. Let the radii of the pins be a and 
B. Describe about a a circle with the radius a sin. (p, or 

a . b 

A —J^ - o . and about b a circle with the radius b ,:;-—- -. 

These may be called the friction circles of the pins. 

If there were no friction, the equal and opposite forces at a and 
B acting on the pin would be in the direction a b. They are really 
in the direction of a common tangent to the two circles. There 
are four positions of this common tangent depending upon the 
direction of relative sliding of pin and piece at each place. 

There are certain cases in which we see the direction without 
trouble. Thus : if a (Fig. 57) is the centre of the crosshead of a 







__^ 


B 

\ 


,-v: 


p 

E 


^0- 


""Y^ 






X 




' 5 










C 



Fig. 57. 



steam-engine, and c of the crank shaft. If the resultant force due 
to the piston-rod, etc., is p at a, with no fiiction, we know that r 
sec. B A c is the pushing force in the connecting-rod a b, and the 
moment of this about c is f sec. bag x c d cos. BAC = rxcD. If, then, 
we had no friction at the slide, but only in the pins a and b, as we 
see that both frictions diminish the turning moment, the line of 
resistance touches the two friction circles drawn at a and b in the 
manner shown in the figure, and the turning moment on the crank 
shaft is F X c E. Happily, in steam-engines that are well attended to 
the fiiction circles are so small that we may practically neglect 
them, and consider the force between a rod and pin to be truly 
through the centre of the pin. Should we, however, as in the case 
of the hinged joint of an arch, fear friction at the joint, it is 
worth while remembering that whatever be the loads on a piece, 



IH 



APPLIED MECHANICS. 




the necessary equilibrating- forces at the pins are to be found 
tangential to the friction circles there. Thus, if a and b (Fig. 58) 
are the centres of the hinges of the arch a c b, draw the friction 
circles at a and b. It is not enough to find the line of resistance 

A B, but also other pos- 
sible ones not passing 
through the centres a and b, 
but touching the friction 
circles there. If a is the 
inclination of the centre line 
of the arch at a, and ;• is the 
radius of the fiiction circle 
there, there will be no great 
error in adopting this r ule ; — 
The line of resistance may 
start from any point between k' and x" in the vertical through a, 
if a'a = a a^ = ?Vsin- a- 

The smaller, therefore, the radius of the cylindric sui-face or 
hinge of a hinged arch the better. When an arch ling abuts at a 
cylindiic surface of large radius, I cannot see much distinction 
between it and an arch fixed at the ends, except that there may be 
no tension in the joint, or rather that the line of resistance may 
not pa5s outside the joint. 

92. Body turning about an Axis. — In Fig. 59 we have a body 
which can move about an axis. It is acted on by a number of 

cords exerting forces 
which just balance 
one another. Now, 
if you make this ex- 
periment you will 
find that you must 
keep your finger on 
the body, because it 
is in such a state 
that a very small 
motion either way 
causes the forces to 
no longer balance. 
Suppose, however, 
you were to let the 
cord A be wound on 
a pulley whose radius 
is equal to the distance A O ; the cord b on a pulley whose 
radius is equal to b o, and so on, you would have the arrange- 
ment shown in Fig. 60, which differs from Fig. 59 in that a 
'small motion has no effect on the balance. Now what is the 




Fig. 59. 



APPLIED MECHANICS. 



115 




Fig. 60. 



condition of balance in this case? Suppose one complete turn 

given to the axis, every cord shortens or lengthens bj a 

distance equal to the 

circumference of the 

pulley on which it 

is wound. Let a and 

B lengthen, and let 

C shorten, then we 

know that the work 

done by A and b 

must be equal to the 

work done against (• 

Hence, 

Pull in A X cir- 
cumference of 

a's pulley, 

together with 

pull in B X 

circumference 

of b's pulley, must be equal to pull in c x circumference 

of c's pulley. 
We might, however, use the diameters or radii of the pulleys, 
and so we see that in Fig. 60 there is balance if 

Pull in A X a 0, together with pull in b x b o, equals pull in 

c X CO. 
The pull in A X A is really the tendency of A to turn the 
body about the axis, and in books on mechanics it is called the 
moment of the force in a about the axis o. The law is then, 
if a number of forces try to turn a body and are just able to 
balance one another, the sum of the moments of the forces 
tending to turn the body against the hands of a watch must be 
equal to the sum of the moments of the forces tending to turn 
the body with the hands of a watch. We sometimes say : — 
"The algebraic sum of the moments of all the forces is zeroT 
That is, we regard one kind of moment as positive and the 
other as negative. Another way of putting the proof is this : — 
If a much magnified drawing be made showing a very small 
motion through the angle 0, it will be seen that the work done 
by c being, either c x motion of point of application resolved 
in direction of c, or motion of point x resolved part of c 
in direction of motion ; in either case this work is equal to 
C X O c X 0, that is, moment of c X 0. Hence, total work = 



116 APPLIED MECHANICS. 

algebraic sum of raoments X 0, and if total work is 0, then the 
ilgebraic sum of the moments is 0. When work is done upon 
A body in turning it, observe that the work in /oot-2?ounds is 
equal to the moment in 'pound-feet multiplied by the angle in 
radians turned through. If a constant moment M acts upon a 
body rotating at a radians per second, Ma is the work done 
per second. If n is the number of revolutions per minute, 
2 TT n is the angle per minute, and hence m 2 tt 7i -f- 33,000 is 
the horse-power. 

[No doubt the student has already become quite familiar 
with the idea expressed by " work = force x distance." It 
has just been shown how, by the very simplest transformation, 
this expression becomes^ in the case where a force is producing 
turning about an axis, " work done == moment X angle turned 
through in radians." Although not required for the study of 
the portion of the subject now under consideration, yet we 
may remind the student that at a later stage it will be an 
-advantage to him if he has accustomed himself to calculating 
work done when the turning moment (called the torque) and 
angle turned through are given.] 

93. The Lever. — Thus, for example, a lever is a body such 
as I have spoken about, capable of turning about an axis. You 
will find that our general rule of work and this rule of 
moments will give the same result. If two forces act on a 
lever, they will balance when their moments about the axis are 
equal; that is, when p, multiplied by the shortest distance 
from the fulcrum or axis to the line in which P acts, is equal 
to w multiplied by the distance of the fulcrum from the line 
in which w acts. 

If a number of forces balance when acting on a lever, the 
sum of the moments tending to turn the lever against the hands 
of a loatch must be equal to the sum of the moments tending to 
turn the lever with the hands of a watch. 

It must be remembered that if the body acted upon has its 
centre of gravity somewhere else than in its axis, then we must 
consider that the weight of the body is a force acting vertically 
through its centre of gravity. 

Uxamjyle. — The safety valve, Fig. 61, must open when -the 
pressure on the valve is just 100 lbs. per square inch. The 
mean area of the valve A, on which we assume that the pres- 
sure acts, is 3 square inches; CD is 2 inches, e is 50 lbs., the 
weight of the lever is 6 lbs., and its centre of gravity is 6 inches 



APPLIED MECHANICS. 



117 



from d; where must e be placed'? All distances are measured 
horizontally. Here, the upward force is 100 x 3, or 300 lbs , 
and its moment about D is 300 x 2, or 600. The moment of 
the weight of the lever is 6 X 6, or 36. The moment of the 
weight E is 50 X the required distance from d. Hence, 600 — 
36, or 564 divided by 50, is the answer; 11-28 inches from 
D. Thus we find where the mark 100 ought to be placed. 

Let the student repeat this for pressures of 90, 80, and 
70 lbs. per square inch, stating in each case the distance of the 
weight from d. What are the distances apart of the marks ? 

This is not the place to consider the various forms of safety 








Fig. 61. 

valve employed by engineers. The force which fluid at rest 
can exert against a valve is not the same as when the valve is 
open and the fluid is moving ; but it is very important that a 
valve should stay open, allowing the fluid to escape. This is 
effected partly by using a peculiar shape of valve, but more 
usually by letting e come closer to the fulcrum when the valve 
opens. In the above figure the valve seat is conical ; in 
practice, a flat seat is now much more commonly used, the 
breadth of the seat being very small. 

Exercise.- — A weighbridg'e consists of three levers whose 
mechanical advantages help each other ; I mean, the short 
arm of each supports the long arm of the next. Suppose that 
the weights of all parts are arranged so as just to be balanced 
when no load is on the bridge, and that the mechanical 
advantages of the three levers are 8, 10 and 12, what load 
will be balanced by a weight of 15 lbs. 1 Ansiver — 14,400 lbs. 
Suppose that it is the first of these levers that is alterable 



118 APPLIED MECHANICS. 

(that is, there is a sliding weight), what is its mechanical 
achantage altered to when the load is 16,000 lbs. ? Answe7' — 
It was 8 : it now becomes increased in the proportion of 16,000 
to U,406, so that it becomes 8-8889. 

Show that the graduation of the lever %vith the sliding 
weight is in equal divisions for equal alterations of the load. 
Do this by finding the position of the sliding weight for 
various loads. 

Friction is greatly got rid of in weighing-machines by using 
steel knife edges as fulcrums of levers resting on hard steel or 
agate plates. Students must examine actual specimens. The 
common chemical balance must be examined. There is one 
thing about it which may trouble the student. " Why are 
short-armed balances now preferred to the older forms ?" The 
short-armed balance has less moment of inertia, and this 
causes it to be quicker in its motions, so that time is saved. 
(See Art. 453.) But there is much more to be said about it 
than this. Indeed, in this as in all other cases, to thoroughly 
understand one machine requires a knowledge of the whole of 
applied mechanics and applied physics, I am not now dis- 
cussing any one machine exhaustively. I was strongly tempted 
to take up the thorough consideration of one machine and call 
this the study of applied mechanics ; and if I had a student 
with a particular interest in one machine this would he the 
very best way to put before him the study of applied mechanics. 
The method I have adopted in this book is to illustrate each 
principle by means of a machine in which that principle 
happens to appear most important. The defect of the method 
arises from its causing a student to think that he knows all 
about a machine when he only knows the most important 
principle of applied mechanics which is illustrated by it. The 
cure for this academic training defect comes when a student is 
compelled to take a special interest in some one machine, and 
't is then, in practical work, that he really is learning applied 
nechanics. We can only partially correct the defect by our 
Qumerical exercise and laboratory work and experience in the 
workshop. 

Students looking at weighing-machines ought particularly 
to notice that when objects to be weighed are placed not on 
swinging scale-pans, but on fairly firm platforms, the construc- 
tion of the balance must be such as to make the measurement to 
be independent of the position of the object upon the platform. 



Applied Mechanics. 119 

EXEECISES. 

1. A uniform straight bar, 2 feet long, weighs 5 lbs. ; it is used as a 
xever, and an 8 lbs. weight is suspended at one end ; find the position of 
the fulcrum where there is equilibrium. Ans., 4-5 inches. 

2. A lever safety-valve has the following dimensions : — Mean diameter 
of valve, 3 inches ; weight of valve, 8 lbs. ; distance of centre of valve 
from fulcrum, 2^ inches ; weight of lever, 1 6 lbs. ; distance of its centre of 
gravity from fulcrum, 13 inches. Find where a weight of 35 lbs. must 
be hung from the lever, so that the steam may blow off at 85 lbs. per 
square inch. Ans., 36-4 inches fi'om fulcrum. 

3. AB CD is a rectangle, and Ac a diagonal. In ac take a point o, 

such that A is a'third of a c. Forces of 30, 10, and 5 act from a to b, c 

to B, and D to c respectively. If a b = 3 inches, and b c = 4 inches, 

write down the moment about o of each force, with its proper sign, and 

find their algebraic sum. . ,. „. 40 20 

Ans.,-iO', +20; +— ; - — • 

4. The diameter of the safety-valve of a steam-boiler is 4 inches ; 
the weight on the lever is 90 lbs., and distance from centre of the valve 
to the fulcrum is 2-5 inches ; what must be the distance of the point of 
suspension of the weight from the fulcrum in order that the valve will 
just lift when the pressm-e of steam in the boiler is 80 lbs. per square 
inch? Length of lever, 30 inches; weight, 12 lbs. ; distance of centre of 
gravity from fulcrum, 14 inches. Ans., 26 '059 inches. 

5. A ball weighing 4 lbs. is fixed to one end of a bar hanging vertical, 
the other end of which can turn about a fixed axis. If the ball be pulled 
out by a force which acts horizontally through the centre of the ball, 
what will be the amount of this force necessary to keep the bar at rest in 
a position such that the bar makes 30° with the vertical? If the length 
of the bar be 30 inches, what would be the force if applied at a point on 
the bar 10 inches from the centre of the ball ? ^4 _ 

^w«-, 3^3; 2^3. 



120 



CHAPTER VIL 




ELEMENTARY AVALYTICAL AND GRAPHICAL ilETHODS. 

94. Ix Fig. 62 we have another example of the fact that when 
there is a displacement of a point, o, in the direction, o t, and 
a force, P, acts in the direction o A, the angle a o Q being ciilled 
a, the work done is the displacement multiplied hv the com- 
ponent of P in the direction o T. This component is P cos. a, or 
it may be defined as the projection of o a, the line which 
represents the force, upon o T. 

The student must note what we mean by the projection of 

a line. The projection of 
^--' o A upon the line o t is o Q. 

If the arrow-head is not 
put upon o Q, the order of 
the letters will indicate the 
sense of the action. Thus 
I shall use Q o to mean a 
sense opposite to o Q. Xow note that if we have lines, 
o A, A B, B c, c D, and D E (Fig. 63), what I call the sum 
of their projections upon any hne is really the projection of 
OK If I have a closed polygon, the sum of the projections 
of its sides upon any line is zero, if the sense of every side is 
circuital round the polygon with the sense of all the rest. 
If the sides are parallel to, and proportional to forces, this 
means that the sum of all the components of all the forces in 
any direction whatsoever is zero. 

Now suppose we have a body acted upon by any number of 

forces in all sorts of directions 
(we can only illustrate this 
by strings and weights) ; and 
if the weight of every portion 
of the body itself be con- 
sidered also acting : and if 
the body is in equilibrium, 
the principle of work tells 
us that if the body receives 
disj)lacement, s, in any direcdon 




Fig. 63. 

a smaU translationai 



APPLIED MECHANICS. 121 

whatsoever (translation means that every point moves parallel 
to the motion of every other point, and through the same 
distance), then the total work done by all the forces is zero. 
But the work done by any force is its component in the direc- 
tion of the motion multiplied by the displacement ; hence, we 
see that the sum of the components in any direction whatsoever 
must be zero. This is a condition which must hold when any 
body is in equilibrium. But it is evident that we have the 
same rule when we say that the forces are proportional to the 
sides of a closed polygon, the senses of the' forces being cir- 
cuital round the polygon. If the forces are not in one plane, 
the polygon will be what is called a gauche polygon, but if the 
amounts and directions of the forces are given in plan and 
elevation, the polygon may be drawn in plan and elevation. 
To give the analytical rule more simply : — Take three lines 
mutually at right angles to one another, o x, o Y, o z ; project 
all the forces upon o x, they balance (that is, the algebraic sum of 
their projections is zero) ; project all the forces upon o y, they 
balance ; project all the forces upon o z, they balance. These 
three analytical conditions are the same as the graphical rule, 
" the force polygon is closed." 

Perhaps, however, we had better confine our attention to 
forces all in one plane. The analytical rule is :— .The algebraic 
sum of all the horizontal components is zero, or the horizontal 
components balance; the algebraic sum of all the vertical 
components is zero, or the vertical components balance. 
These two conditions are the same as "the force polygon is 
closed." 

EXERCISE. 

Griven the following forces, find their equilibrant — 
Force o x of 50 lbs., force o A of 20 lbs., the angle, x o A, 
being 33°; force M o of 56 lbs., the angle XOM being 150°; 
force G of 100 lbs., the angle x o g (all angles are measured 
counter-clockwise) being 217° ; force h o of 70 lbs., the 
angle x o H being 315°. Now let a student draw these on 
paper. I have assumed them all to be in lines through a 
point o, but this was for ease in setting the question. 
The forces need not act through one point. Let him, 
by drawing, or by using a table of sines and cosines, find 
the components of all in the direction o x and o Y, if 
o Y is perpendicular to o x (Art. 31). I make these projec- 
tions to be — 



122 



APPLIED MECHANICS. 



In the direction ox. 




In the direction oy 


50 COS. = 50 




50 sin. = 


20 COS. 33° = 16-773 




20 sin. 33° = 10-893 


+ 56 COS. 30° = 48-497 




- 56 sin. 30° = - 28 


- 100 COS. 37°= - 79-863 




- 100 sin. 37°= - 60181 


— 70 COS. 45° = - 49-497 




+ 70 sin. 45° = 49*497 


Algebraic snni = 2 x — - 14-09 


Al 


^ebraic sum = 2 y = — 27 791 



The Greek letter 2 is used to mean " the sum of all such 
terms as." x means any p cos a, and y any p sin a. We have 
found 2x and 2y ; call them x and y respectively. Draw o c 
== - 14"09, B= - 27-79, and complete the rectangle o b r o ; 
then o R is the resultant and r o the equilibrant. It is 

evident that o r- = \/x^ + y2=z31*16. Also tan r o c=y/x = 
1-971 ; therefore roc = 63° 7', and x o r=:243° 7'. 

The student must get accustomed to symbols which so 
greatly shorten our use of language. 
Our rule is: — If p^, Po, etc., are 
forces in a plane, and if they make 
angles a^ oo, etc., with any line, say 
a horizontal line, then Pj cos. oj = 
Xp Pg cos. ao = Xo, etc., are the 
horizontal components of the forces. 
The directions of arrow-heads must 
be noted, and the algebraic sum of 
the horizontal components is de- 
noted by x = 2p cos. a, or 2x. In 
the same way y= 2p sin. a, or Sy, 
denotes the algebraic sum of the 
vertical components y-^ = Pj sin. a^ 
Yo^Pg sin. a^, etc. Then the re- 
Fig. 63a. ^ sultant, R, of all the forces is the 
resultant of x horizontally and y vertically, and r^ = x" + y", 
and if r makes the angle a, with the horizontal, then tan. a = 
Y -^ X. We have only found the amount, clinure and sense 
of R. We do not yet know where it acts. (The word clinure 
was, I believe, invented by the late Prof. Thomson ; the word 
direction implies more than what we try to express by clinure.) 




APPLIED MECHANICS. 123 

Similarly, if the forces are not in the same islane, and if each p 
makes the angles a, 13, and y, with Ihree standard lines at i-ight 
angles to one another (usnally called the axes of x, i/, and s), then 
if p cos. a is called x and 2p cos. a = x, if p cos. /3 is called y and 
2p cos. y3 = y, and if p cos. y = z and 2p cos. 7 ="z, then r^ —-^2 
H- X- + Y^ and the cosines of the angles which r makes with 
the axes are x/j^, y/j^^, and z/^. 



95. Any physical quantit}^ which is directional is called a 
vector (such as a displacement, a velocity, an acceleration, a 
force, a stress, the flow of a fluid, etc.), and may be represented 
by a straight line. The length of the line represents the quantity 
to some scale of measurement ; the line's clinure represents 
the clinure of the vector, and the barb of an arrow represents 
its sense. It is easy by actually drawing lines and measuring 
their lengths to solve problems which would otherwise require 
a good deal of mathematical knowledge. This sort of graphi- 
cal calculation having proved useful, it has attracted the 
attention of men who have leisure enough to make an elaborate 
study of its methods. It has, unfortunately, become a com- 
plicated weapon with which these men can attack all sorts of 
problems which are much more easily solved in other wa?ys. 

We shall only use graphical methods where they happen to 
be the best methods. Now a force is a vector ; it has mao:ni- 

o 

tude, clinure, and sense, but it is more than a vector ; it has 
a fourth quality not possessed by <^rdinary vectors — namely, 
actual position in space. Settling any one point in its line of 
application settles its position when its clinure is known. But 
if we are told that forces all act at a point, they are added 
exactly as all mere vector quantities are added. When, then, 
I speak of finding the resultant or equilibrant of forces at 
a point, I may be said to speak of any vectors. 

96. Forces acting at a Point. — The line a b (Fig. 
represents a force in clinure by its own clinure, 
in amount by its length to any scale we please, 
and in sense by its arrow-head, which shows 
that the action of the force is from A to b. 

It would not be correct to call this the force 
B A, because this is opposed to the sense of the 
arrow-head. The forces A o, o b, o c, o d 
(Fig. Go), all act upon a small body, 0, or their 
lines of action when produced all pass through -pig. 64. 




124 



APPLIED MECHANICS. 




a point, 0, in a large rigid body. The amount of each force 
is sho%vn by the length of the line, representing it to some 
scale. iSTow to add these forces together in the most perfect 
manner — that is, to find a force called their resultant, which 
shall be exactly equivalent in its effects 
/ to all the above forces acting together 
— we draw a polygon (Fig. 66). ^Each 
side of this polygon is parallel to, and 
proportional to a force in Fig. 65 ; 
thus, the side a corresponds to the force 
A o, and the arrow-heads agree, and 
lastly the action indicated by the 
arrow-heads is circuital. Fig. ^Q is 
always called the force polygon. 
When it is unclosed, as it is in the 
present case, we know that the forces 
A 0, etc. (Fig. 65), are not in equili- 
brium. To keep A o, o b, etc., in equilibrium, a new force, 
called the equilibrant, must be introduced corresponding to 
the side e (shown dotted), which will close the polygon, its 
arrow-head being circuital with the others. Now if we 
want the resultant of a o, o b, c, o D, it evidently acts 
through and corresponds to e, Fig. 66, but with arrow- 
head reversed. The resultant of a number of forces is equal 
and opposite to their equilihrant. 

Prove now the following statements by actual drawing : — 
1st. The resultant of any number of forces 
does not depend on the order in wliich they 
are drawn as sides of the polygon. 

2nd. Any lines or forces whatever which 
form a closed polygon in any given order will 
form a closed polygon if drawn in any other 
order. 

3rd. In adding forces we may first find 

the resultant of some of the forces, and then 

add together this resultant and all the rest of the forces. The 

answer will always be the same, however we may group ti.e 

forces before adding them. 

When the forces are not all in one plane the polygon must 
be drawn by descriptive geometry, and to draw it, and so find 
the resultant or equilibrant as the closing side, is an excellent 
graphical exercise. 



Fig. 65. 




APPLIED MECHANICS. 125 

In "working exercises we recollect the fact that the resultant 
of all the forces due to gravity is called the weight of the body 
and acts through its centre of gravity (seepage 136). This one 
force replaces the millions which are due to gravity. When 
we observe that we have only tliree forces acting upon a body 
at rest, we know that they must be in a plane and act through 
a point unless they are parallel. In the ordinary books on 
mechanics there are many problems which are easily solved 
if we remember this fact. Four forces in equilibrium, if not 
all in one plane, must meet in one point. When a body touches 
a smooth surface we know that the force which there acts upon 
the body is normal to the smooth surface. When a body 
touches a rough surface we know that the force which there 
acts upon the body makes an angle with the normal, and the 
limiting value of this angle when sliding is about to occur is 
what is called the angle of repose, or the angle whose tangent 
is /i, the co-efficient of friction there. It is astonishing what 
a number of exercises are easily worked out if one will only 
recollect these few general principles. 

The student will at this place work again the exercises of 
page 111. It ought to be getting clear to him that the most 
difficult analytical work has really nothing in it more com- 
plicated than these exercises have. The exercises are usually 
called easy, certainly students get easily into the way of 
working them, and I am always sorry to notice too great 
an ease of this kind. It often indicates shallowness of 
comprehension. 

EXEECISES. 

1. Forces o a = 30 lbs,, o b = 50 Ihs., c o = 15 Ihs., d o = 80 lbs., 
E = 150 lbs. ; the angles are b o a = 45°, c o a = 90°, d o a = 135°, 
E o A = 270°. Find the resultant analytically and graphically. 

Ans., 223 lbs. at an angle of 303" with oa. 

2. Sheer legs each 50 feet long, 30 feet apart on horizontal ground, 
meet at point c, Avhich is 45 feet vertically above the ground ; stay fro]]:i 
C is inclined at 40° to the horizontal; load of 10 tons hanging from c. 
Find the force in each leg and in stay. Ans.^ 7"8 tons ; 64 tons. 

97. In many engineering problems, when forces A, b, c, d, 
etc., are given, it is sometimes important to be able to 
show graphically the resultant of A and b, the resultant of A, b, 
and c, the resultant of A, b, c, d, and so on. Thus (Fig. 
67) A, b, c, d, etc., are given forces. 

Draw the unclosed force polygon (Fig. 68). Join the point 
with each corner of the force polygon. From the point b' 



126 



APPLIED MECHANICS. 



where A and B meet (Fig. 67) draw a line b' c' parallel to the 
line o B c (Fig. 68) (o b c is the line from o to the corner 
where b and c meet) ; from c' draw c' d' parallel to o c D, 
and so on. Then b' c' represents the position and direction, 




and b c represents to scale the resultant of the given forces 
A and B. Similarly d'e' represents the position and direction, 
and ODE represents to scale the resultant of the given forces 
A, B, c, and D. Kote the arrow-heads of the resultants we 
have found. The line a b' c' d' e' p' (Fig. 67) is usually called 
a line of resistance. 

98. We have in Art. 93 confined our attention to the forces 
acting upon a small body, or forces which all pass through one 
poini if they act on a large body. But in 
¥ig. 67 and in our description we assumed a 
large body, and our forces were any forces 
whatsoever. TS'e gave it a small motion of 
translation, and obtained an important j-esult 
from the consideration that, on the whole, no 
work was done upon the body. ZsTow, let us 
assume that any point o in the body is fixed, 
or, rather, that an axis o is fixed, the axis 
being at right angles to the plane in which 
all the forces act ; about this axis we assume 
that the body may rotate. Consider the work 
done by all the forces during any small 
rotation d : it is zero. But the work done 
by any force is, as we have already seen 
(Art. 92) the moment of the force multiplied 
by 0. Hence the sum of all the moments of all the forces about 
is zero if there is equilibrium. But any small plane motion 




Fig. 68, 



APPLIED MECHANICS. 127 

of the body whatsoever we know to be resolvable into a 
motion of translation and a rotation about some axis o at 
right angles to tiie plane. Hence the law of work tells us 
that if any system of forces is in -equilibrium their compo- 
nents in any direction balance one another and their 
moments about any axis balance one another. 

In the numerical exercise Art. 94 let the respective forces 
(only given in amount, clinure, and sense as yet) be at the 
following distances from a certain point, which I shall call s. 
The 'sign + means that a force tends to turn the body against 
the hands of a watch, ox is at the distance + 5 feet, OA 
is at — 2 feet, mo is at — 7 feet, o g is at -|- 3 feet, h o 
is at — 4 feet. Let the student now draw these forces in their 
proper positions relatively to s. The sum of their moments 
is 50 X 5 - 20 X 2 - 56 X 7 + 100 X 3 - 70 X 4, or 
— 162 pound-feet. This is the moment of their resultant 
which is 31-158 lbs. ; so that its distance from s is — 5*2 feet. 

99. If the cliuurc and sense of a force p be given, it is also neces- 
sary to give some point through which its direction passes. Thus, 
let all the forces "be in one j)lane ; let r make an angle a with the 
horizontal ; let the co-ordinates of the given point be x and y 
referred to the axes of x and y. 

If p COS. a is X and p sin, a. is v, then as the moment of p 
about any axis is equal to the sum of the moments of x and y, 
taking moments about the origin, the mojnent of p is y a? - xy. 

If ii is the resultant, its components are 2 x and 5 y ; call them 
X and Y. Also, if x and y are the co-ordinates of a point in r, 
XY — ^ x = 2 (yk — Xl/). 

For equilibrium we must have 2 x = . . (1), 2 y = . . (2), 
2 (y « - X ?/) = . . (3). 

Notice that we may have (1) and (2) true without (3) being 
true. In this case the sy.stem of forces reduces to a mere couple 
whose moment about any point is 2 (y a? — x y) ; such a system is 
caUed a torque. Jf we choose any point in the plane, we can 
replace any system of forces by a single force through this j)oint, 
together with a couple whose moment is the sum of the moments 
of the system of forces about this point. This is often an 
exceedingly important fact to remember. (See Art. 100.) The 
student ought to work many numerical exercises graphically and 
analytically. 

Example 1. — A beam, abode, is supported at a and e by 
forces X and y. The load at b is 3 tons, and a b = 4 feet ; load at 
c is 2 tons, and a c is 7 feet ; load at d is 1\ tons, and a d is 9 feet ; 
AE is 12 feet. 

Here ^-fy = 3 -f- 1 -\-1\-=^1\ tons. Taking moments about 
A, the moments with the hands of a watch are 

3x4-t-2x7 + 2ix9 = 48i ton-feet. 



128 



APPLIED MECflA\ICS. 



1 


^ ^x 


Tz 


i 


I 






Fig. 69. 


Q. 



The moment against the hands of a watch is y x 12, and x 
has no moment, because it acts at a. Hence our second equation 
is 12 y = 48§, u = 4*0417 tons, and therefore x is 3"4o83 tons. 

Example 2. — V^e neglected the weight of the beam itself in 
Example 1. If its centre of gravity is 4 feet from a, and if the 
weight of the beam is half a ton, and if x' and y' are the additional 
supporting forces, 

^ -^y' = h I X 4 = / X 12. 
Hence, ^ = I ton, x = \ ton. 

Example 3. — In Fig. 69 o A is part of a beam. Considering 
only the portion of beam to the 
right of the section at o, let the 
loads downward, Pj, Pg, and I's, 
and the supporting force upward, 
Q, be given. Let the perpendicular 
distances from o be Aj, a.-,, a^, and 
q. Find a force through o, and 
a couple to balance the given 
forces. [Call the force s. If it 
acts downward at o, its amoimt 
must be s = Q — p^ — 'Pg — v^. 
If the couple is called m and it 
acts tending to turn the body o a round o, with the hands of a watch 

M = Qg - Pj «! — P2 «2 - P3 flg. 

When we come to consider beams we shall call s the shearing force 
and M the bending moment at the section o. 

The student will at this point work the Exercises 1, 2, 3, of pages 
134-0, as weU as the following : — 

1. A uniform beam, 20 feet long and supported at its ends, has weights 
of 1, 3, 2, and 4 cwts. placed at distances of 2, 8, 12, and 15 feet respect- 
ively from one end. Taking the weight of the beam to be 5 cwts., find 
the reactions at each of the supports. Ans., 7 cwts. ; 8 cwts. 

2. Draw anv Line x, and lines o p, o q, o e, o s, o x, making angles of 
28=, 62=, 118=, 220=, and 305° respectively with ox. Consider thatlorces 
act along these Knes, their amounts being 25, 34, 14, 42, and 18 lbs. Find 
the amount and direction of the force which balances these. In doing 
this first determine the components of each force in the directions o x and 
o Y, and arrange these in columns as shown in Art. 94. 

A)is., 15-67 lbs ; 232°-2 with o x. 

3. A trap- door of uniform thickness, 5 feet long and 3 feet wide, and 
weighing 5 cwts., is held open at angle of 35° with the horizontal by 
means of a chain. One end of the chain is fixed to a hook placed 4 feet 
vertically over the middle point of the edge on which the hinges are, the 
other end being fixed to the middle point of the opposite edge. Deter- 
mine the force in the chain and the force at each hinge. 

Afis., 2-65 cwts. ; 2*5 cwts. 

4. A uniform beam, weighing 2 cwts., is suspended by means of two 
chains fastened one at each end of the beam. When the beam is at rest it 
is found that the chains make angles of 100° and 115= with the beam; 
find the tensions in the chains. Ans., 1 cwt. ; 1-1 cwt. 



APPLIED MECHANICS. 129 

5. A B is a iLOrizontal uniform bar 1^ feet long, and r a point in it 10 
inches from a. Suppose that a b is a lever tliat turns on a fulcrum under 
F, and carries a weight of 40 lbs. at b ; weight of lever, 4 lbs. If it is kept 
horizontal by a fixed pin above the rod, 7 inches from r and 3 inches from 
A, find the pressure on the fulcrum and on the fixed pin. 

Ans., 89-14 lbs. ; 45-14 lbs. 

100. When the forces are not in one plane, let the force p make 
angles a, jS, and y with the three axes of co-ordinates. Let a point 
in the direction of p be x, y^ z. If p cos. a, or x, p cos. )8, or y, and 
p cos. 7, or z be the three components of p, we can use x, y, and z 
instead of p for aU purposes. Positive directions are the direc- 
tions of increasing x, y or z. Thus the moment of p about any 
axis is equal to the sum of the moments of x, y, and z about the 
axis. Attention must be paid to the sense of each force, and 
whether it tends to turn the body against or with the hands of a 
clock. The student ought to spend time in fixing clearly in his 
mind the truth of the following statements : — The moment of p 
about the axis of a? is z «/ — y 2 ; the moment of p about the axis 
of 2/ is X 2; — z a? ; the moment of p about the axis of z is y a? — Jiy. 
Hence we see that if b, is the resultant, its components are S x, 
2 Y, S z, and the sum of their squares is s?, which is therefore 
easily calculated ; also each of them, di^dded by r, is a direction 
cosine of r. Again, if a?, y^ and z be the co-ordinates of a point 
in B, then 

^2z — «SY=S(zy - Y^), 

2 2 x - ^ S z = 2 (x 2 — z a?), 

»2y - ^2x==S(Ya?- xt/). 

The value of b and its cUnure (the angles which it makes with the 
axes) having ali-eady been found, these equations enable the posi- 
tion of a point in the resultant to be found ; so that r is completely 
determined. 

For equilibrium we must have 2x = o, 2y==o, 2z = o, 
2 (z y — Y z) = o, 2 (x 2 - z a?) = o, 2 (y a? — x y) = o. 

Given a set of forces, it is evident that we can always sum them 
into a resultant force acting at any point we please to choose, 
together with a couple about some axis. If we are not given the 
point, it is always possible to reduce any system of forces to a 
resultant, and a couple whose axis is the resultant force. 

101. The Link Polygon. — We shall now consider graphical 
methods of dealing with forces which do not necessarily act 
through one point. Take, for example, the forces of 1, 2, 3, 4 
(Fig. 70). Draw the unclosed force polygon 1', 2', 3', 4' (Fig. 
71), with its sides parallel to and proportional to the forces, and 
the arrow-heads circuital. Now the dotted line h a, with its 
arrow non-circuital with the rest, is parallel to and proportional 
to the resultant of all the given forces. But this does not tell 
us where the resultant force is situated, although it tells iis its 
direction and amount. 



130 



APPLIED MECHAyiCS. 



From any point, o (Fig. 71), dra^r a line to the junction of 
1' and 2' (it is easier to say draw the line o 1' 2'), o 2' 3', etc., 
to all the angles of the force polygon. Xow construct a new 
unclosed polygon, with its corneii on 1, 2, 3, 4 (Fig. 70), and 
its sides parallel to o 1' 2', o 2' 3' etc. (Fig. 71), its last side 
being parallel to oa, and its fii-st parallel to oh. V^e have 
now found the point, 5 (Fig. 70), where the first and last 
sides of the link polygon' meet The resultant of the forces 





Fig. 70. 



Fig. 71. 



1, 2, 3, 4, 



5, and corresponds to 
mamitude. The new 



,, J., passes through this point, 
the closing side, ha, in direction and 
polygon is called the link polygon of the forces relative to 
the pole, 0. The position of a, the point at which we stait to 
draw the link polygon, may be chosen anywhere on 1, and hence 
there may be any number we please of link polygons for a 
given position of the pole, o. Again, there are any number 
we please of link polygons corresponding to any other posi- 
tion of 0, and we can choose o where we please. Any student 
who studies this in the light of what he did in Art. 96, will 
see that the link polygon really consists of a system of links 
which wotdd be in equilibrium under the given set of forces 
and the force we have found ; and since the mere links only 
introduce forces which are of themselves in balance, being 
equal and opposite in each link, the system of forces acting at 
the joints must balance. 

Suppose we find that wben we are given the forces 1, 2, 3, 
and 4 'Fig. 72), and we draw the force polygon 'Fig. 73}, and 
any link polygon Fig. I'l , that these are both closed, let ns prove 
that the forces are in equilibrium. 

A system of forces acting on a rigid body is not affected by 
introducing any number C'f forces which separately balance one 
another. Xow let a force represented by the length of the line o 
1' 2' act at the point a in the direction b a. its sense being shown 
"b^f the arrow-head near a, and let an equal force act at b in the 



APPLIED MECHANICS. 



131 



direction ab, its sense being- opposite to that of the force at a. 
These two forces are in equilibrium with one another, and they 
cannot therefore affect the original system of forces in any way. 





Pig. 73. 

Similarly, the forces shown by the arrow-heads in b c, c d, da are 
introduced, every pair balancing one another. 

Now we see that the three forces at the point a are in equi- 
librium with one another, because they are parallel to and pro- 
portional in amount to the sides of the triangle omn (Fig. 73), 
and corresponding arrow-heads would run right round the triangle. 
Similarly, there is equilibrium at every other corner of the link 
polygon A B c D ; hence all the forces are in equilibrium, and hence 
the forces, 1, 2, 3, 4, taken by themselves, must be in equilibrium. 

The theorems which we wish students to prove by construction 
can be proved to be generally true, reasoning from the fact that a 
number of forces acting at a point can only have one resultant. 

102. We see, then, that the force polygon alone is sufficient 
to find the resultant of any number of forces if the forces meet at 
a point, but we need also the link polygon if the forces do not 
meet at a point. 

The Unlc polygon really shows that the sum of the turning- 
moments of the forces 1, 2, 3, 4 (Fig. 70) about any point is equal 
to the moment of the resultant about the same point. The force 
polygon pays no regard to tm^ning moments of forces ; it merely 
tells us about the resultant of the forces, supposing- that they all 
pass( d through the same point. 

103. You ought by actual drawing to illustrate the truth 
of the following four ways of putting one statement. If you use 
coloured inks your drawings will be more instructive. 

1st. The resultant of any number of forces is independent 
of the order in which we draw them in the force polygon, and 
draw between them the sides of the link polygon. 

2nd. In adding forces we may first find the resultant of 
some of the forces, and then add together this resultant and all 
the other forces. The answer will always be the same, however 
we may group the forces before adding them. 



132 APPLIED MECHANICS. 

3rd. If the force polygon of a number of forces is closed, 
and if we can draw a closed link polygon, then all the link 
polygons we may draw will also be closed. 

4th. If any other pole be taken in Fig. 71, and another link 
polygon be drawn and a new point 5 (Fig. 70) is found, both of 
the points so found lie in a straight line parallel to 6 a of Fig. 71. 

Yoii will also find, and it is easy to prove, that the locus of the 
point in which any two sides of the link polygon meet is parallel 
to the ]ine which closes the corresponding portion of the force 
polygon. Again, if J is taken as pole instead of o, the last side of 
the link polygon is found to he in the direction of the resultant of 
the forces 1, 2, 3, 4; and, generally, any side of the link polygon 
is in the direction of the resultant of the corresponding numher of 
the given forces. Thus, if b is taken as pole, 4, 5 hecomes the 
resultant of the forces 1, 2, 3, 4, and 3 4 hecomes the resultant of 
the forces 1, 2, 3. It is evident from this that the direction of the 
resultant of any two forces, or of any numher of forces, which meet 
at a point passes through their point of intersection. 

A system of forces may not reduce to a resultant force, hut he 
equivalent to a couple. When this is the case the force polygon is 
closed, and the first and last sides of any link polygon that may he 
drawn are parallel to one another. You may also find it worth 
your while to prove hy construction this statement : if two link 
polygons are drawn for two positions of the pole o, the correspond- 
ing sides of the two polygons meet in points which lie in a 
straight line parallel to the line joining the two positions of the 
pole o. 

If you have heen ahle to make a few drawings such as I have 
heen speaking ahout, and so take an interest in this easy and 
instructive method of working mechanical exercises, you ought to 
work hy means of it some such exercises as the following : — 

104. 1. Exercises. — In Exercise 2 of page 128, draw the force 
polygon, taking the forces in the order op, or, o q, o t, o s, and 
observe that the resultant and equilihrant are the same as before. Obtain 
also the resultant of o p, o r, o t, and the resultant of o a, o s, and show 
that these have a resultant equal to the resultant of the five forces. 

2. Draw a line h k 47 inches long. On h k take points a, b, c, distant 
from H 0-4 inch, 1-5 inch, and 3-3 inches respectively. Now di-aw a p, b q, 
c R incbned at angles of 72°, 57°, and 37° with h k. Suppose that a force 
of 214 lbs. acts from k to h, one of 576 lbs. from r to c, one of 132 lbs. fi'om 
Q to B, and one of 237 lbs. from p to a. (1) Draw a force polygon to deter- 
mine the amount, clinure, and sense of the resultant. (2) Take any pole 
o and draw a link polygon to determine the lateral position of the result- 
ant. In giving your answer say what are the perpendicular distances of 
H and K foom the resultant. (3) Draw a new force 23olygon, taking the 
forces in a different order, and observe that the resultant is the same as 
before as regards amount, clinure, and sense. With respect to a new 
pole 0, now draw a second link polygon, and observe that the lateral 
position of the resultant agrees with that obtained before, (4) Take the 
first force polygon and choose a new pole, and with respect to it draw a 



APPLIED MECHANICS. 133 

new link polygon. Observe that this gives rise to the same resultant. 
That is, the closing sides of each link polygon meet at points, all of which 
should lie on the same straight line. (5) Calculate the above . answers 
according to the rules of Art. 99. 

Ans., (1) 1,066 lbs.; 39 7° with hk; sense same as force rc; (2) 
1-312 inches; 1-7 inches. 

105. Exaiwple. — Given a set of forces, find two forces, one 
given in position, and clinure the other to pass through a given 
point p, such that these two forces will balance the given set. 
The ingenious student will find out how to do this himself; 
other students will benefit by the following instructions. 

Draw the force polygon with one corner unknown ; choose 
a pole and begin drawing a link polygon from the point p ; the 
side closing the link polygon enables the missing corner of the 
force polygon to be found. This problem is one of the very 
commonest to come before the engineering student. Thus, let 
any structure (a roof principal or a railway girder, for example) 
have given loads. Let it be supported at a hinge joint at p, 
and upon rollers at Q, to allow for expansion. The direction 
of the supporting force at Q is known, and one point, p, in the 
other supporting force is known. 

106. A student who sees the essential ideas underlying our 
methods of working will have pleasure in working curious 
problems, such as the following exercise : — Given a set of 
forces and given three points, A^ b, and c. Draw a link polygon 
with three of its sides passing respectively through a, b and c. 

Hint. — You must first find the resultant of the three given forces, and 
observe where its line of action meets y y ; this point of intersection, i, 
joined to x, gives the line of action of the balancing force through x. 
Three forces now act at i. Their lines of action are known, and the 
magnitude of one force ; hence the amounts and senses of the other two 
can be found. 

107. Exercise. — Draw a rectangle abcd, ab = 5 inches, b c = 1-8 
inch. From a, along a b, measure off lengths, a e, a r, a g= 1-75, 2-8, 5*7 
inches respectively. From n along u c measure off n h, n k, n l=: 1-4, 2-4, 
2"85 inches respectively. Suppose that forces of the following amounts act 
on a rigid body — viz. , 1,460 lbs. from a to h, 1,085 lbs. from e to k, 808 lbs. 
from r to L. These are balanced by two others, one of which acts through 
X, a point in d a distance 0*6 inch from n, and the other has for its line of 
action y y, the line passing through c and g. Find the magnitudes of the 
balancing forces and the angle between them. (The rectangle is intro- 
duced merely as a convenient way of settling, without a figure, the given 
forces.) Scale \ inch to 100 lbs. Ans., 2,550 lbs. ; 1,140 lbs. ; 60°. 

108. Example. — Given a set of forces, a, b, c, d ; given also 
three lines, x, y, z, as the positions of three forces which are to 
balance our given set. Find these three. The method her© 



134 



APPLIED MECHANICS. 



is not so obvious as the method of the last example. Draw 
a', b', c', d', known sides of the force polygon. Draw, also, 
after d', x' parallel to x unlimited in length. In it choose o 
the pole and join to all known corners of the force polygon. 
Note that o x' is itself two radiating lines from o, because there 
are two corners of the force polygon in it. Now let the point 
where d and x meet be called a side of the hnk polygon ; the 
intercept on x till it meets y is another side, and it will now 
be found that we have sufficient information for the completion 
of both force and link polygons. As an example, Fig, 74 
represents a ladder whose centre of gravity is at G, and weight 
300 lbs. A string fastened to it at c in the direction c o keeps 
it in equilibrium, its end A resting on the smooth wall A, and 
its end b on the smooth floor o b. Find the pull in the string 
and the reactions at a and b. 

The forces acting on the ladder are shown by the arrows. 
DraAv Y x (Fig. 75) vertically to represent the weight of the 
ladder. Draw w Y horizontally of unknown length. Draw 



A _ 










P 


\ 






/ / 


'' 




\ 


<.. 


'-fw 


'■■ 


^^^ 


^ 


■'c\ 


X 















B 





Fig. 74, Fig. 75. 

X parallel to o c, and take o anywhere in this line. Use o as 
pole of the force polygon. Join o y. Now the link polygon 
is M N P M, and drawdng w parallel to P m, and w z parallel to 
b m, we find that Y x z w is the force polygon. The lengths of 
X z, z w, w Y represent the forces at c, at b, and at a. 

The student will find it very instructive now to introduce 
friction into this problem, and thus create two new problems. 
(1) If the pull in the cord is just sufficient to prevent the 
ladder from moving. Here the angle o B M ought to be 
90° — 0j where tan. 0^ is the coefficient of friction at B, and 
o A P ought to be 90° + 0^ if tan. <po is the coefficient of friction 
at A. (2) If the pull in the cord is just sufficient to produce 
motion. In this case, o b m = 90° + ^^ o A p = 90° — tft^. 



APPLIED MECHANICS. 135 

Students ought at this stage to work a great num ber of 
exercises. 

EXEECISES. 

1. In Fig. 74, A B = 20 feet, o b = 10 feet, b c = 2 feet. The weight of the 
ladder is 300 lbs. Find the pull in c o and the reactions at b and a. If /* = 
0- 1 , find the pnll in the rope ; first, if b is made to approach o ; second, if b is 
ahout to recede from o. Ans., 99-21'hs. ; 3191bs.; 98 lbs. ; liolbs.; 58-8 lbs. 

2. A horizontal beam a f has loads, at b of 1 ,000 lbs., at c of 250 lbs., at 
D of 1,200 lbs., at E of 600 lbs. If a b = 5 feet, a c = 9 feet, a d=15 feet, 
AE=18 feet, af = 24 feet, find the su^jporting forces analytically and 
graphically. Ans., 1,548 lbs.; 1,502 lbs. nearly. 

3. Draw a rectangle ab c d, ab = 5'2 inches, and b c = 2-2 inches. On 
AB take points e, p, g, h, k, l distant from a, 0-9, 1-55, 2-7, 3-25, 4*25, 
and 4-75 inches respectively. On bc take bm = 1-25 inch. On cd take 
N, o, p, Q, distant from c, 1'5, 2-85, 3-7, and 4 inches respectively. On 
D A take D K, = 1 inch. Now, three forces act on a rigid body — one of 
0-615 ton fi'om f to o, one of 0536 ton from h to p, one of 0*423 ton from 
L to N. These three forces are balanced by three others, whose lines of 
action are, x x drawn through r and g, y y drawn through e and a, and 
z z drawn through m and k. Find the magnitudes of these balancing 
forces. Scale i inch to 0-1 ton. Ans., 0-930 ton ; 0-375 ton ; 0-815 ton. 

109. The distance x of the centre of mass (usually called cen- 
tre of gravity, but only few bodies have true centres of gravity) 
of a body or systera of bodies from a plane is obtained by 
multiplying each little portion of mass m by the distance x 
of its centre from the plane, adding together, and dividing by 
the whole mass. The symbol for this is x . m = ^ m . x where 
M stands for ^ m. Practically the engineer often finds the 
centre of gravity by an experimental method. 

The distance x of the centre of an area (usually called the 
centre of gravity of the area) from a plane (or more usually of 
a plane area from a line in its plane) is obtained by multiplying 
each little portion of the area a by the distance x of its centre 
from the plane (or line), adding together, and dividing by the 
whole area. The symbolic way of representing this is a? . a = 
^ ax, where A stands for the whole area. 

If the centres of the masses m■^^, m.2, m.^, etc, are at the dis- 
tances Xi, x^, x.^, etc., from any plane, let the sum of the masses 
m-y -f m2 + etc. be called m, and let x = {m^ x^ + m.^ x.j + etc.) ~ m, 
or, as we prefer to write it, a? ^= S {mx) -h- m. Similarly, taking 
distances from two other planes at right angles to the first, let 
y='2.{my) -j- m, 2 = S (m z) -^ m. Let x, y, z be the co-ordinates 
of a point. If u-^, u^, M3, etc., be the distances of the masses from 
any other plane, at a distance a from the origin, perpendicular to 
the direction (J, m, n), it is easy to see that 



136 APPLIKD MECHANICS. 

and if M is the distance of the point alre^y fonnd from the new 
plane, u = Ix -\- mt/ -\- n z — a: and using the above values for 
X, y, z, and re-arranging terms, we find that u = '2 [m u) -~ m. 
We see,, then, that the above-mentioned point will be in the same 
position whatever be the planes of reference. We call it the centre 
of mass or centre of inertia; and when we are dealing with a 
system so small that the forces of attraction upon it due to gravity 
may be regarded as parallel to one another, we may call it the 
centre of gravity. In the same way, if the centres of the areas 
<?j, a~2. etc., all in the same plane, are at the distances Xj, x.2, etc., 
fi-om a line in the plane, and if a is <?i — ^^ -r etc., the whole area, 
and if x is the distance of the common centre of area (often called 
the centre of gravity of the area) from the line, then 

X = (a-j x-^ + flg ^2 + 6^^-) "^ ^• 

110. We can use a graphical statics method of finding the 
centre of mass, or area g. Thus, let there be masses or areas. 
7/2-,, 771.2, etc., whose centres are at the points 1, 2, etc. (Fig. 76). 
Draw parallel forces 11, 22, et-c,, in any dii-eetion proportional 
to 7nj, '//2.1, etc., and find the resultant by the above method. 
Suppose M y to be the direction of the resultant. Now 
repeat the process, taking the parallel forces of the same 
magnitudes as before, but in a different direction, and let 

MP be the direction of the resultant. Evidently 

M, where these lines meet, is the centre of 

^.^ \fX^\ gravity of the masses or areas. This method 

^V *ikl\j ^^y often prove useful, for areas especially. 

<J3 yfe)''r\ Thus, to find the centre of gravity of any 

\' \ \\ \ given area, divide it into any suitable number 

'p \ \ \ \ \ ^^ parts, so that the centre of gravity and area 

\ \ \\ \ of each part, may be found easily. If we 

■• ""' " \n divide the area by parallel lines, these lines 

Fig. 76. may t-e drawn equidistant, and the area of 

each part is approximately given by the length 

of the line which separates it from either of its neighbours. 

A repetition of the process has been employed to determine 

the moment of inertia of the area about any given line. 

111. I do not advise students to adopt this link polygon 
method of finding centres of gi-avity or of calculating moment 
of inertia. A practical engineer will always apply the ordinary 
formula to find the centre of gravity of an area. Thus, if 
you want the centre of gravity of the figui'e M N P (Fig. 77), 
draw two parallel lines, g H, k o, touching the figui^e at two 
opposite sides. Draw a line, K G, at right angles to G H, and 



APPLIED MECHANICS. 



137 



divide it into any number of parts, each equal to d. Draw 
the lines a B, c d, etc., parallel to g h, so that they are at the 
distance d apart, the distance from A b to G H, or from Y z to 
K o being J d. It is evident that if 
N X P is a line parallel to G H through 
the centre of gravity, then approxi- 
mately 



G X = i<? . 



ab + 3cd + 5ef + etc. 





^-^ ~~^~- 




/.r zx^ 




/ \ 


/ 




"/ 


X f! 


/ 


7 




/ 




1 




/ 


'\ 


.y 







Fig. 77. 



AB + CD + EF + etc. 

We have thus obtained one line 
through the centre of gravity, and in 
a similar way we may obtain another 
such line, and their point of intersec- 
tion is the centre of gravity required. 

Sometimes we choose as our line 
of reference, a line, G h, which cuts 
through the area ; in this case distances on one side of the line 
are to be considered negative ; and if g H happens to pass 
through the centre of gravity, of course the sum ^ a a? is 0. 

We often cut the shape of an irregular area from sheet 
zinc and balance it in two positions on a straight edge to find 
the centre of gravity of the area. 

Exercises. — 1. Masses whose centres are in a straight line at a, b, c, d 
are 4 lbs., 8 lbs., 7 lbs., 6 lbs. ; where is the common centre of mass if 
A B = 0-5 feet, A c = 2 feet, and a d = 1\ feet ? Ans., a g = 1-32 ft. 

2. A disc 8 inches diameter, 2 inches thick, with a hole 4 inches 

5" 





<-2"-- 








( _... 6" * 






7 


2.8i 


5^:.J(? 


^ 












Fig. 78. 



diameter ; centre of hole, 1 inch from centre of disc ; where is the centre 
of mass ? Ans. , ^ inch fi-om centre. 

3. ABC is an equilateral triangle, each side being 3 raches long; 
particles whose masses are 1, 2, 3 are placed at a, b, c respectively; find 
their centre of gravity by construction, and note its distance from a. 

Am.f 2-18 inches. 



^ 



138 APPLIED MECHANICS. 

4. A B c D is a square lamina of uniform density ; e, f are the middle 
points of A B and b c. If the corner of the square is turned down along 
the line e f, so that b comes on to the diagonal a c, find the centre of 
gravity of the lamina under the new circumstances. 

Ans., -^-g of the diagonal from the centre. 

5. A circular disc, 8 inches in diameter, has a hole 2 inches in 
diameter punched out of it, the centre of the hole being 3 inches from the 
circumference of the disc. Find the centre of gravity of the remaining 
portion. Ans., 0-0667 inch from centre of larger circle. 

6. A thin plate of metal is in the shape of a square and equilateral 
triangle, having one side common; the side of the square is 12 inches 
long. Find the centre of gravity of the plate. 

Ans., 2-86 inches from centre of square. 

7. Find the centres of area of the areas in Fig. 78. 

112. To fi.nd the moment of inertia, i, of a great numher of little 
masses about an axis, multiply each mass by the square of its 
distance from the axis, and add up. If the whole mass is m, we 
often write uk^ = i; and when i and m are known we can 
calculate k, which we call the radius of gyration of the mass about 
the axis. To find what is called the moment of inertia, i, of a great 
number of little areas about a line in their plane, multiply each by 
the square of its distance from the line, and add ujj. If a is the 
whole area, and a A^ = i, we call k the radius of gyration of the 
area about the line. 

113. Just as we found centres of graAdty, so we may obtain the 
moment of inertia, i , of any area about any line ; or, as is often 
the case, suppose we wish to find the moment of inertia of m n o p 
(Fig. 77) about nxp, a line which jmsses through the centre of 
gTavity. Evidently the moment of inertia about g h is approximately 
i = ^=^(ab + 9cd + 25ef + etc.)/4. 

Students ought to practise this method first upon a rectangle 
and a circle whose moments of inertia have been worked out for 
them by the calculus. 

Now, it is well known that the moment of inertia of an area 
about any line is equal to its moment of inertia about a parallel line 
through its centre of gravity, together ivith the product of the area 
into the square of the distance between the two lines. Hence, the 
moment of inertia about NXPisio = i — gx2.<;(ab + cd + ef 
+ etc.). . 

It will be found in practice that this easy way of carrying out 
simple ideas is better than the compli- 
cated use of the link polygon method 
— N for finding moment of inertia. If the 
area may be divided into rectangles 
whose sides are parallel to and per- 
pendicular to the axis, we need not 
subdivide these rectangles. It must be 
remembered that the moment of inertia 

of any given area such as a rectangle 

° '^' about any axis is equal to the area 

^ig. 79 multiplied by the square of the dis- 

tance of its centre of gravity from the 



D ^ 



APPLIED MECHANICS. 139 

axis, plus the moment of inertia of tlie area about a parallel line 
through its centre of gravity. Thus the rectangle a b c d 
(Fig. 79) is known to hare, about n m n, the moment of inertia 

— — — '■ so that the moment of inertia of the rectangle about 

o' o o' is 

A B . B C^ . „ /B C2 \ 
- +AB.BC.M02, orAB-Bcfy^- +M02j 

It is mainly by the use of this rule that we have found the 
moments of inertia of the various sections shown in Table VI., and 
all these ought to be worked out as exercises by students. 

The student will find it good exercise to take a few sections of 
angle-iron, T-iron, rails, and other specimens of rolled iron, and 
find by the above graphical method the position of centre of 
gravity of each section, and the moment of inertia of each area 
about any line through the centre of gravity. The exact forms 
ought to be taken from real specimens. If the area is symmetrical, 
one Hne through the centre of gravity can always be found by 
mere insi}ection. 

In using this or any other gTaj)hical method, it is well to know 
what is the error involved in having each strip of width d, 
instead of being iafinitely narrower. We ought to add to the sum 
in (2) the moment of every strip about its own central line — that 
is, # (ab + CD + EF, etc.)/i2 or Ad^/-^^, if a represents the total 
area. It is easy to see that in a rectangle of dej)th d, if we divide 
into strips of breadth d, the fractional error is d^/D^. 

114. When the moments of inertia of an area about any three axes 
through a point are known, the moment of inertia about any other 
axis through the same point may be found ; because if a distance 
be measured from the point along an axis which is equal to the 
reciprocal of the radius of gyi^ation of the area about the axis, the 
extremities of all such measured distances lie in an ellipse. The 
principal axes of the area are in the directions of the major and 
minor axes of this ellipse. Thus, if for any area, m n p R (Fig. 80), 

the least moment of inertia is about an axis, o a, and is — -^, and if 

the greatest moment of inertia is about ob, and is — x, then 

ob2' 

ab a'b' being an ellipse whose major and minor axes are A a' and 

B b', the moment of inertia about an axis, o o, is — s. 

oc* 

115. To know the principal moments of inertia of an area is 
important for many purposes. It is specially important in regard to 
struts. A strut will bend ia such a way, that the axis through the 
centre of gra^dty of a section, at right angles to the plane of bend- 
ing, is the axis about which there is least moment of inertia of the 
section. The ellij)se lets us see the moments about all axes. To draw 
it for any particular section, if the section is symmetrical, as in sections 
of Fig. 80, we know that the axis of symmetry and the axis at right 
angles to it are the principal axes of inertia. If the section is not 



uo 



APPLIED MECHANICS. 



symmetrical, as in the case of an angle-iron, we find the moment 
of inertia about three axes as o p, o q, o n of Fig. 81, and we set 
off the distances o p, o q, o r to 
represent the reciprocals of the radii 
of gyration. "We now have the 
graphical problem : given three 
points p Q R of an ellipse, and its 
centre o to draw the ellipse. Mr 
Harrison thinks the following solu 
tion better than any other. 

With centre o, radius o q, de 

scribe a circle, and through h 

where p r cuts o q, draw the chord 

pr such that p b. : n r = f n. : Kii 

Draw the radius os perpendicular 

^^' to o a ; through s draw s m, « n 

parallel respectively to ^ o, or; and* through n and m draw lines parallel 

to R o, o p, faitersecting in s ; then o s is conjugate to o q. 





Fig. 81. 

For proof, suppose the ellij)se orthogonally projected into a 
circle, and let the smaller figure be similar to this projection. 
This figure can be drr,wn, remembering that parallel lines project 
into parallel lines the mutual ratios of whose lengths remain 



APPLIED MECHANICS. 



141 



unaltered ; also that conjugate diameters project into perpendicular 
diameters. The solution given consists in drawing this figure 
with oq coinciding with o Q, and then locating s. 

To determine the j)rincipal axes, through s draw a line (not 
shown) perpendicular to o q, and on it take two points d, e, 
opposite ways from s, such that sd = se = oq. Then the axes 
of the ellipse are resiDCctively equal to the siim and difference of 
o D and 3, and the major axis bisects the angle doe. 

MOMENT OF INERTIA OF A RECTANGLE. 

116. The moment of inertia of a rectangle about the line o o 
through its centre, parallel to one side. — Let a b 
= b, yic = d. Consider the strip of area between 
p = y and o Q, = y + Sy. Its area is b . 5?/, 
and its moment of inertia about o o is ^ . t/^ . 8t/ ; 
so that the moment of inertia of the whole 
rectano-le is 



i 



-\d 



y^ . dy ov b 






b^. 
12 




The moment of inertia of the area a b c about 
an axis o at right angles to the area is equal to 
the sum of the moments of inertia Xg and i2/, about 
o X and o y, axes at right angles to one another Fig- 82. 

in the area. For if a is a portion of area at p, ix 
is the sum of all such terms as os . p e,^, \y is the sum of all such terms 
as « . p a^ ; and the sum of each such pair of terms is a term a . o p^. 
117. Moment of inertia of a circle about its centre.— Consider 

the ring of area between the 
radii r and r + 5r. Its 
area is lirr . Sr more and 
more nearly as Sr is made 
smaller and smaller, and its 
moment of inertia is 2 Trr^ M. 
The integral of this is 
•IttR"* for a circle of radius 
R. The square of the radius 
of gyration is \v?. Now, 
in this case \x = ly, each 
being half of ^ttr^ ; so that 
the moment of inertia of a 
circle about its diameter 
is Ittr^, or e^TTD^ if d is the 
Fig. 83. diameter. 

118. The student ought 
to be able to prove the propositions referred to in Art. 112 : — 

1. As to mass or inertia. — To prove that the moment of inertia 
about any axis is equal to the moment of inertia about a parallel 
axis through the centre of gravity together with the whole mass 
multiplied by the square of the distance between the two axes. 
Thus, let the plane of the paper be at right angles to the axes. 




U2 



APPLIED MECHANICS. 



Let there be a little mass m at p in the plane of the paper. Let o 
be the axis through the centre of gravity of the whcle mass, and 
o' be the other axis. We "want the sum of all such terms as 
ii..(o'p)2. 

K^ow, (o'p)2 = (o' o)- + OP- + 2 . o o' . o Q, Tvhere q is the 
foot of a perpendicular from p upon oo', the plane containing 
the two axes. Then, calling 2 m . (o' p)^ by the name i, calling 
2 m . o p^ by the name ig, the moment of inertia about the axis o 
through the centre of gravity of the whole mass, then 

I =^ (o' o]^ 2 m -f i^ + 2 . o o' . 2 m . o Q. 
But 2 ffi . o Q means that each portion of mass m is multiplied by 
its distance from a plane at right angles to the paper through the 

centre of gravity, and this is zero. So that 
(^ __,p the proposition is proved. Or, letting 2 «» be 

called M, the whole mass, 

1=1, + M.(o'o)». 
2. To prove the proposition about areas. 

Let o o (Fig. 85 ) be the axis through the 

centre of gravity, and o' o' a parallel axis. 

"We want i, the sum of all such terms as a 
0^/ (o' p)-, and this is the same as 2 « . o p^ -f 

2 2.' . o p . o o'+ 2a . o o'-. Bat 
Fig. 84. 2 2<3r.op.oo' = 2.oo'2a.op, 

and this is from, our definition of centre of gravity ; so that 

I = lo 4- o o' 2 a. 





EXEECISES. 

1. A fly-wheel has a rim of rectangular section, the outside and insidf 
radii being 8 feet and 7 feet, "What is the err-rr in assitming the radiui 
of gyration to be 7 "5 feet? 

^ns., The true radius of gyration 
is 7"516 feet, and hence the assumed 
radius of gyration is -21 per cent. 
wrong. It would give a moment of - 
inertia '4 per cent, wrong. 

2. Find the radii of gyration of 
the sphere of Table n.,p. 2-51, al>:)ut a 
line touching its surface ; the solid 
cylinder about a line touching its ~ 
outside, parallel to the axis ; the rod 
about a Ime at right angles to it at one 
end. Ans.. -.5916 <?/ olAJ; -5773 f. 

3. "What error \& introduced in Table H., by neglecting the size of the 
secHon of the prismatic rod \ 

4. Two homogeneous spheres of weights 12 and 20 lbs., radii 0-2 and 
0*3 feet, their centres 5 feet apart : find the distance of g, the centre of 
gravity, from o, the centre cf the smaller ; find i^. the moment of inertia 
about an axis through g at right angles to o g ; find the moment of inertia 
about o G itself ; find the moment of inertia about any line g a if the 
an?le ogais a. ^/w.. 37o inches ; 27131 : 131-3 : 131-328 + 27000 sin.^ a 



Pig. So. 



APPLIED MECHANICS. 143 

119. Given the moments of inertia of a lamina about a pair of 
axes at riglit angles in tlie jilane of the lamina, to find the moment of 
inertia about any other axis thi'ough the intersection and lying in 
the plane. Let the moment of inertia ahont o x he ix and the 
moment of inertia about o y be ly ; o p is any other axis, the 
angle x o p being a. Let the distance of a small portion of area 
at a from the three axes be called y', x', and y ; then 

y = p' cos. a — x' sin, a, 
and 7^ = yf2 QQg_2 ^ _|_ ^'2 gjj^_2 a — 2 x' if sin. a cos. o. 
Hence, if i is the moment of inertia of the whole area about the 
new axis, i = la: cos.- o. + ly sin.^ a — lixy sin. a cos. o . . . . (1), 
where ixy is written to mean the product of inertia about the axes 
X and y, or the sum of all such tenns as " portion of area x .ry." 

Now, let I = -4-' la; = 4r, 13/=—, ^xy = 4- 



r- 1 



r,2' 



When A is the area, r, r^^ and r-^ are the reciprocals of radii of gyra- 
tion ; but s has no name. 

If we take a point p in the 
line p, such that o p = r, and let 
the co-ordinates of this point 
relatiyely to the original axes be 
X and y^ then (1) becomes 
aj2 2/^ Ixy _ . 

which is the equation to an 
ellipse. We see, then, that if a 
distance proportional to the re- 
ciprocal of the radius of gyration Fig. 86. 
about an axis be measured along 

that axis from o, the points so found lie in an ellipse. When the point 
o is the centre of the area this ellipse is called the momental ellipse 
of the area, and its principal axes are the principal axes of inertia. 
If they had been chosen as the axes of reference, evidently L^y, the 
product of inertia relatively to them., would have been zero. 




APPENDIX TO CHAPTER YII. 

Mr. J. Harrison, of the Royal College of Science, has been 
kind enough to prepare the following short account of the 
general principles involved in the graphical study of forces 
when they do not act in one plane, and readers are referred to 
"Graphics," by Prof. R. H. Smith, for more detailed information. 
Forces in space and framed structures in three dimensions. 

Problem 1. — To find the resultant of a given system of 
forces in space^ whose lines of action all pass through a point. 

A force in space is conveniently defined by two orthogonal 
projections of the line which represents it. These projections 



144 



APPLIED MECHANICS. 



represent the resolved parts of the force respectively parallel 
to the planes of projection. 

To compound the given system, add the forces as vectors. 
Two projections of the gauche polygon representing the vector 
addition are required. 

Let p Q R, p' q' r' (Fig. 87), be the given plans and eleva- 
tions of the lines of action of the forces. 

The plan p q r, and the elevation 2^' 4 ^' of ^^ vector 
polygon can be at once drawn, since the lengths of the sides 
are supposed given as part of the data. Then a line, s, s', 




Fig. 87. 

through the given point o, o', parallel and equal to the closing 
side s, s' of the vector polygon, represents the required resultant. 
Problem 2. — Tlie lines of action of a concurrent system of 
forces in equilib7'ium in sjmce being given, and the magnitudes 
of all the forces except three^ to find the three magnitudes. 



APPLIED MECHANICS. 



145 



In Fig. 87 the forces p p', q q', r r', are those given 
completely ; the magnitudes of l l', m m', n n', are required. 

Compound the known forces into a single force s s', by 
means of the vector polygon shown in plan and elevation. 

To find L l', observe that its magnitude must be such that 
the component perpendicular to the plane of m n shall be equal 
to the component of s s', perpendicular to the same plane. 

Draw a new elevation on a plane perpendicular to the 
plane of M n. Let s" or o" a" be the new elevation of s. Draw 
a" b" parallel to m" n", to meet l" in b" ; then o" b" is the 
magnitude of l", the elevation of L on x y . Project b" to b. 
Then o b is the length of tlie plan l. 

Draw I parallel and equal to o b, and close the vector 
polygon in plan by drawing the lines m, n respectively parallel 




.^^ 



Fig. 88. 

to M and N. The elevation of the polygon can now be drawn. 
The true lengths (not shown) of the lines I V, m m', n n would 
give the actual magnitudes of the three forces, l, m, n. 

Problem 3. — To reduce two given forces^ acting in directions 
at right angles to each other, to a single force and a couple, the 
'plane of the couple being perpendicular to the line of action of 
the single force. 

Let p and q (Fig. 88) be the given forces, and let A b of 
length a be the line meeting both p and Q at right angles. 

At B introduce the equal and opposite forces p^ and Pg asi 
shown, in a line parallel to P and equal to it in amount. 



146 



APPLIED MECHANICS. 




APPLIED MECHANICS. 147 

Compound p^ and Q into R^. In the plane A b r^ draw a d, 
making, with a b, the angle dab=:qbRj = 0; and draw 
B D perpendicular to A D. Introduce the forces r and Rg, each 
equal and parallel to R^ as shown. 

Then the required single force equivalent to p and Q is R, 
and the required couple r x c d. 

For let the couple p Po be represented by the axis b n = p x 6t; 
resolve this into b m, m n, where m n is perpendicular to B M. 

Then b m = b n x cos. — p « cos. 6 = r a sin. d cos. d = 

RXC D. 

And M N = B N X sin. S = T a sin. d -- n a sin.^ = r x B C. 

So R X c D represents the component couple b m, and r, RoUiay 
be taken as the couple represented by m n. Now Ro cancels Rj, 
and there remain r and R x c d. 

The two outer lines at c d may be taken to represent the 
couple. The convention adopted as to sign is, that when the 
arrow-heads on the couple axis, and on the force line point the 
same way, as in the figure, the tendency of the forces is to 
produce a right-handed screw motion, and conversely. 

Problem 4. — To com^oound a given general system of forces 
in space. 

Let the lines of action of the forces be supposed cut by any 
plane, and at the points of intersection resolve the forces 
respectively along and perpendicular to the plane. Compound 
each of these two sets. The system is thus reduced to tioo 
non-intersecting forces at right angles. If desired, these two 
forces may be compounded, as in the last problem. 

The given system illustrated in Fig. 89 consists of three 
forces, p, Q, R, the projections of the lines of action of which 
on three planes mutually perpendicular are shown, as are also 
the three projections of the vector polygon, drawn as if the 
given system were concurrent. 

The system may be reduced to two forces — one in the 
horizontal plane x y, the other vertical. 

To find the former, draw a Knk polygon in plan with 
respect to any pole o. Thus, s = s is the horizontai force. 

To find the vertical force, two methods are available. 

(1) Compound the vertical components. In the figure, the 
elevations of the components on z x are shown, and their 
resultant v' is found by means of a link polygon drawn with 
respect to any pole o^. In a similar manner (not shown), the 
elevation on y z of the resultant vertical force could be found. 



148 APPLIED MECHANICS. 

thus completely determining the vertical force and giving the 
plan H of its line of action. 

(2) Or thus : — On z x, by means of a link polygon drawn 
to any pole o', find s', the elevation of the line of action of the 
component force of the system parallel to the plane z x. And 
similarly, on y z, find s", the elevation of the component parallel 
to Y z. Let these intersect the ground lines respectively in li 
and A", projections from which give h, the plan of the line of 
action of the vertical force required. Its magnitude is a^ d'. 

A further construction is shown for reducing the forces, as 
in the last problem. The line of action of the single force thus 
obtained is called the central axis of the system. The couple 
is equal to the moment of system about the central axis, and 
may be shown to be the minimum couple. There is no other 
axis about which the forces have a less moment. 

]\^ofe. — If the given system contain couples as well as 
forces, treat those independently, adding their axes as vectors. 
Resolve this couple parallel and perpendicular to the central 
axis, and add to the other part of the system. 

Problem 5. — To determine the stresses in a non-redundant 
framed structure of three dimensions under given loads. 

The criterion for a non-redundant stiff frame in three dimen- 
sions, with non-rigid joints, is that the number of bars must 
be six less than treble the number of joints. For this relation 
is evidently true in the simplest example — viz., for a frame of 
six bars forming a pyramid ; and in building up a frame which 
shall be stiff, each new joint requires three new bars. If any 
portion consist of a plane frame with redundant members (such, 
for example, as a plane quadrilateral with crossed diagonals), 
such redundant members are not to be counted in applying the 
criterion. 

Fig. 90 shows a derrick crane carried by a braced triangular 
pier. 

The stress diagram for the frame is built up in plan and 
elevation, applying Problem 2 in succession to the several 
joints. The order of taking the joints is that indicated by 
the Roman numerals, and the various points on the stress 
diagram are marked a, b, c, etc., in order as they are found. 

If the pier had been braced as shown in Fig. 91, then after 
having drawn the stress diagram for the joints I and II, 
there would be no new joint with less than tour unknowns. 
In this case the stress in the bar marked P could be found by 



APPLIED MECHANICS. 



149 




Fig. f 1. 



150 APPLIED MECHANICS. 

resolving the forces at the joint V perpendicular to the plane 
which contains the other three unknowns, Q, R, s. Tlie building 
up of the diagram would then proceed in the same order as 
before. 

EXERCISES. 

1 . A rertical crane post is 10 feet high., jib 30 feet long, stay 24 feet long, 
meeting at a point c. There are two back stays making angles of 45° 
with the horizontal ; they are in planes due north and due west from the 
post. A weight of 5 tons hangs from c. Find the forces in the jib and 
stays — 1st, when c is south-east of the post; 2nd, when c is due east; 
3rd, when c is due south. 

2. A tripod whose vertex is a, and whose legs are a b, a c, a d, of lengths 
8, 8-5, and 9 feet respectively, sustaias a load of 2 tons. The ends b, c, d 
form a triangle whose sides are b c = 7 feet, cd = 6 feet, b d = 8 feet, find 
by graphical construction the compressive forces in each leg. 

Ans., 1-16 ton, 0-55 ton, 0*53 ton. 

3. Three ro]Des, each 12 feet long, hang from the three corners of a 
horizontal isosceles triangle, a b c, in which a b and a c are each 20 feet, 
and B c is 10 feet. The ropes are joined at their ends and support a load 
of 1 ton. "J^'ind the pull on each rope. Ans., 0-52 ton, 0-52 ton, 0-92 ton. 



151 



CHAPTER YIII. 



EXAMPLES IN GRAPHICAL STATICS. 

120. In any structure, such as the principal of a roof and 
many girders of bridges formed of many different bars, if we 
neglect the weights of, or upon, the bars, and we assume that 
the joints are frictionless hinges, it is easy to see that the force 
exerted by any bar is in a straight line between the centres of 
the hinges at its ends. For whatever may be the many forces 
between a piece and pin, at the surface of the pin, these forces 
must all be normal to the surface, since there is no friction ; 
they must therefore all be directed in radial lines, and hence 
their resultant must be a radial line through the centre of the 
pin. In this case, then, the joints of a structure only being 
loaded with known forces, it is easy to calculate the pushing 
or pulling force exerted by each piece. 

To illustrate this — let there be three pieces, whose centre- 
lines are A o, bo, and c o (Fig. 92), 
meeting on a pin whose centre is o. Sup- 
pose we know that the piece c o pulls the 
pin in the direction c Avith a force of 
2,000 lbs. We have then to find the two 
forces in the given directions of A o and 
B to balance the known force o c. 

Draw the triangle (Fig. 93), whose 
sides, G, a, b, are parallel to o c, A o, and 
B ; and let c represent the force o C 
or 2,000 lbs. to some scale, and let the 
arrow-head on c represent the sense of o C. 
Then the lengths of h and a represent the 
other two forces to scale. Put the circuital 
arrow-heads on h and a, and we see that 
o A is a pulling force, and the piece o a is called a tie -rod ; b o 
is a pushing force, and the piece b o is called a strut. Note that 
B is attached to some other pin than o. If we study the equi- 
librium of the new pin we must remember that b o pushes it, and 
we must draw the new arrow-head when we study the neic pin. 

It is very important that the student should illustrate an 
important fact like this for himself in the laboratory. Fig. 94 
shows two real pieces A o and b o hinged at o. Hang on any 




Fig. 93. 



152 



APPLIED MECHANICS. 



weight c. Adjust the screw at a' until the spring-balance a' a 
is in a line with o A, so that it indicates the pull on o A. The 
push in B is recorded by the spring-balance b' b. When 
things are at rest, open the parts of a two-foot rule until they 




Fig. 94. 

just fit the angle between a o and b o, and transfer this angle 
to a sheet of paper. Now do the same with the angle between 
B and c o, and test on the paper if the sides of a triangle 
parallel to a o, b o, and c o really represent, to some scale, the 
three forces. If they do not, speculate for yourself upon the 
discrepancy ; how much discrepancy is due to the fact that 
the forces as measured are not truly the forces at o, because of 
the weights of the parts ? How much is due to the fact that 
the rule presupposes a pin with absolutely no friction ? If you 
are ingenious and an advanced student, you will try a pin with 
much friction, or perhaps a riveted joint at o, and so learn 
more than I can tell you. 

EXERCISES. 

1 . In Fig. 94 the jib o b makes an angle of 45*^ with the horizontal ; 
ruigle A o B, 16° ; weight at c, 5 tons ; find the forces in o a and o b. 

Ans., 13-66 ton, 16-73 ton. 

2. A chain fastened at o (Fig. 94) goes round a snatch-block at c, 
and then over a pulley at o in the direction o a to the barrel, the total 



APPLIED MECHANICS. 



153 



weight at c being 5 tons ; find tlie forces in o a and o b. Dimensions 
same as in 1, Ans., 10-76 tons; 16'73 tons. 

3. If a wharf crane, the post, tie-rod and jib measure 15, 20, and 30 
feet respectively, what would be the stresses in each of the three members 
when a load of 7 tons is suspended over the pulley at the jib-head (1) 
when the lifting-chain passes from the pulley to the drum parallel with 
the jib : (2) when the drum is placed so that the chain passes from jib-head 
parallel to tie-rod ? 

Ans., 7 tons ; 2-3 tons; 11-67 tons ; 11-2 tons ; 9-3 tons ; 18-67 tons. 

4. A contractor's portable hand-crane has a vertical post a b, to which 
the jib A c is inclined 45°, and the tension -rod b c makes with a b an angle 
A B c of 120°. The back-stay, from the head of the post b to the extremity 
D of the horizontal strut a d, is inclined at an angie of 45° to a i>. Find 
the counterbalance weight required at d to balance a load of 10 tons sus- 
pended from the end c of the jib. Determine also the nature and amount 
of the force in the jib a c, and in the rods b c and b d. (The tension in 
the chain may be neglected.) 

Ans., 23-66 tons ; 33-46 tons ; 27-32 tons ; 33-46 tons. 

121. Let us now consider the roof-principal shown in Fig. 95. 




Fig. 95. 

Certain loads are given acting at the joints, and we know 
that the structure is supported by two forces or reactions at its 
two ends. Our first step is to find these two supporting forces. 
They must be in equilibrium with all the external loads. 

Now, it is well known that we must be given either the 
direction or some other information about one of these support- 
ing forces, else the problem becomes indeterminate. It is 
usual to be told that one or other supporting force is vertical. 
This condition is arrived at in practice by having at one end of 
the structure a shoe with rollers resting on a horizontal 
plate of iron. 

The notation which we use very materially simplifies the 



154 



APPLIED MECHANICS, 



process of calculation. You observe that every space between 
two forces in Fig. 95 is indicated by a letter. The line which 
separates the space A from the space B is called A B, and cor- 
responds with the line A B in Fig. 96. A 2^oint is indicated 
by the letters of the spaces which meet at that point. Thus, 
A G H is the end of the roof-principal. 

Suppose the supporting force at the point f g p is known to 
be vertical. We must first find the amount of the force f g, 
and the direction and amount of the force A G. 

Draw the force polygon, A B c D E F, Fig. 96."^ We see 
that to close it we need two lines to join f and A. Now, one 




Pig. 90, 



of these, F G, is vertical. Take o as pole. Join o A, o b, o c, 
o D, E, F in the usual way. Draw the link polygon, shown 
dotted in Fig. 95, commencing at the only known point of the 
force A G, namely A G ii. Now, o g (Fig. 96) is parallel to the 
last side of it, and thus we find p g and G A, the supporting 
forces at the end of the principal. Having found the two 
supporting forces, F G and G A, we proceed as follows : — We 
have the closed force polygon A b c d E F G A. The arrow-heads 
shown on this force polygon are not to be rubbed out during 
the calculation, and in practice we mark them in ink. All 



* In an actual case there -wonld usually be loads given at the end joints 
as well as the others. In this first example I have only taken the loads as 
shown. 



APPLIED MECHANICS. 155 

other aiTOW-lieads which we draw on Fig. 96 may require to 
be rubbed out, and ought only to be marked in pencil. We 
must begin our calculation at a joint where only two pieces 
meet, and where one force which acts there is given. Now 
at the joint A G H we know the force A g. In Fig. 96 draw 
A H aad G H parallel to the pieces a h and G h of Fig. 95. 
Put arrows on the sides of the triangle G A H circuital with 
the arrow on G A. Now we see by the arrows that the piece 
A H picshes the joint with a force represented to scale by th^j 
length of the line a h (Fig. 96). We know, then, that A h 
is a strut, since it pushes, and we know the total pushing force 
in it. Similarly, H G is a tie, and the total 'pulling force in it 
is represented by the length of the line h g in Fig. 96. 

We now rub out the arrows which we are supposed to have 
drawn in pencil on the lines A h and h g (Fig. 96), and proceed 
to the joint A B i H. It must be remembered that although the 
pieces A h and b i are in the same straight line, we regard 
them as two separate pieces. 

We know the force A B, we also know that the force with 
which the piece A H pushes the joint (we have already found it 
to be a strut, therefore it pushes both joints at its ends) is 
represented by the length of the line a h (Fig. 96). Draw, 
then, HI and bi (Fig. 96) parallel to the pieces H i and bi 
(Fig. 95). We have thus a polygon a e i h. The force a b 
(Fig. 96) tells us how to pencil arrow-heads circuitally 
round this polygon. When we do this we find that the piece 
B I pushes the joint with a force represented by the line b i 
(Fig. 96), so that b i is a strut. Also IH is a strut, in which 
the stress is represented by the line i h (Fig. 96). We pro- 
ceed in this way from joint to joint, always taking care to rub 
out our pencilled arrow-heads when we proceed from one joint 
to the next. The lengths of the lines in Fig. 96 give the 
magnitudes of the forces in the pieces of the structure. It 
is easy to prove that, if no mistake is made, no discrepancy 
will appear when the drawing is being finished. 

If in Fig. 96 the points K and i were found to coincide, 
this evidently means that the piece K i is unnecessary in the 
structure. If, again, we find that we cannot close one of our 
little polygons in Fig. 96, we ought to proceed to new joints, 
and, pDssibly, when we again consider the joint with which we 
had difiiculty, we shall be able to close its polygon. If we 
still find difficulty, it must be caused by two or more joints, 



156 APPLIED MFXHAXICS. 

and the pieces connecting these are evidently unnecessary to 

the structure. If we find in Fig. 96 two points with the same 
letter, we evidently require to add a new piece to the structure, 
which will exert a force represented by the distance betwe<"^n 
these two points. 

]S'o explanation in writing will enable the student to master 
this beautiful method of determining the forces in structures. 
He must select structures, apply loads to the joints, and calcu- 
late the various forces for himself. When he has made four 
such calculations, he vrill know nearly all that can be said on 
the subject. (Figs. 99 and 101 show two examples.) 

122. Roofs. — It is not my object here to describe the con- 
stiiiction of a roof or a bridge. For such information the 
student must examine real structures and good drawings of 
roofs and bridges for himself. 

Suppose, for instance, that he finds a roof, somewhat Like 
his own, to weigh — including possible snow, etc. — 40 lbs. per 
square foot of horizontal are^ covered. Suppose his principals 
are to be placed 8 feet apart, the span being 50 feet, then each 
principal has to support about 

8x50x40, or 16,000 lbs. . 

Now, if Fig. 97 is the shape of his principal, as A B, b c, 

c D, and D E are aU equal, we may suppose that, however the 

roof covering may be sup- 
ported by the principals, the 
piece of rafter, a b, or any 
other of the divisions, sup- 
ports 4,000 lbs. The joint B 
gets half the load on a b and 
half the load on b c : conse- 
quently the load at the joint B is taken to be 4,000 lbs., 
and similarly for c and D. The joints a and E have 2,000 lbs. 
each. 

When the above vertical loads have been given to the joints, 
we have to consider wind pressure on one side of the roof. If 
we suppose, as we reasonably may, that 40 lbs. per scjuare foot 
is the greatest pressure of wind ever likely to occur on a surface 
at right angles to the direction of the wind, then the normal 
pressure per square foot on roofs of the following inclinations 
may be taken from the following table, which is obtained from 
experiment. 




APPLIED MECHANICS. 



167 



Angle of Roof 


5° 


10° 


20° 


30° 


40° 



TABLE I. 

Normal Pressure of Wind against Roofs. 
Angle of Roof. 



Normal 

Pressure. 

6-0 

9-7 

18-1 

26-4 

33-3 



Normal 



50° 
60° 
70° 
80° 
90° 



88-1 
40-0 
40-0 
40-0 
40-0 



Tlius, if the portion of one slant side of the roof between two 
principals has an area of 240 square feet, and if the inclination 
of the roof is 30°, say, then 240 x 26*4, or 6,3361bs., ^ 
has to be supported by each bay. Transferring i\s^ 
this to the joints, we see that at b (Fig 98), in 
case the wind pressure is upon the side A b c, we 
have the vertical load x b, or 4,000 lbs, due to 



weight of roof. 



and also Y b, or one 




Fig. 98. 



half of 6,336 lbs., normal to the roof, and due 
to wind. Complete the parallelogram, and evi- 
dently z B is the load at the joint b which we must use in 
our calculations. 

The student will find that if a roof-principal can only be 
supported by a vertical force at a certain end, the stresses in 
the structure are greatest when the other side of the roof is 
acted on by the wind. 

123. Many joints in a real structure are usually stiff joints, 
so that many pieces may really be subjected to bending, a,s 
well as to direct compressive or tensile 
stresses. A general method of taking 
stiffness of joints into account is quite 
unknown ; but when we discuss bend- 
ing we shall see pretty clearly what is 
the effect of a stiff joint, and in some 
cases we shall be able to make calcula- 
tions on the subject. It may generally 
be assumed that the strength of a 
structure is greater if the joints are 
stiff than if they are merely hinges. 
This is not always the case, and, 
from the indeterminateness of the 
problem of finding the stresses in a 
structure whose joints are stiff, many 
large bridge trusses are made with Pig. loo. 




158 



APPLIED MECHANICS. 



nearly all their joints hinged. In roof-principals the joints 
are often made stiff, rather for the purpose of stiffening the 
whole structure than for the sake of strength. We shall 
presently see the distinction between stiffness and strength 
in structures. In a roof all joints of struts are usually made 
stiff. What we shall now say is of more importance in bridges 
than in roofs. 

If two or more pieces of a structure are in a straight line 
with one another at joints where they meet, it is usual, for 
strength, to make the joints between them quite rigid. Thus, 



L 


Wi 




N 


J. 


» J, ^ , 




::i^ 


/\ 

D 


v 


/\ 

F 




J 


F\'A- 10]. 




Fig. 102. 

the pieces A H and B i of Fig. 95, or a b and B c of Fig. 97 
ought to form one bar. But this is only useful when the 
pieces in question are struts, and our reason for the continuity 
of the pieces is that a strut is stronger when its ends are fixed 
than when its ends are not hxed. Thus the piece B i (Fig. 9fi) 
will resist a greater thrust if it is continuous with a h and c K 
than if it were hinged with these pieces. (See Art. 372.) It is 
not good in all cases to fix the end of a strut by rivets, etiC., 
instead of a hinge ; because the benefi^t due to fixing an end 
may be more than counterbalanced by the evil effects of bending 
introduced to the strut through the joints by a tendency to 
change the angle which the strut makes with the piece to wldch 
it is fixed. The common-sense of the engineer will always 
enable him to decide as to the judiciousness of fixing the end 
of a strut. 



APPLIED MECHANICS. 



159 




124. Sections of Structures. — It is often of considerable im- 
portance to find immediately the forces in pieces of a structure 
which are not near the ends. If we can draw any surface 
which will cut through the 
pieces in question, we can 
calculate the stresses in 
these pieces directly, sup- 
posing the pieces are only- 
three in number. Thus, 
the section ace (Fig. 103) 
cuts the pieces b A, B C, and 
D E. Now, not only the 
whole structure, but every part of it is kept in equilibrium. 
What forces keep the part A h E in equilibrium 1 They are 
the known forces at b, f, and h, together with three unknown 
forces Avhose directions are b A a', b c c', and d e e'. Given 
the directions of three forces which equilibrate a number 
of known forces, we know that they may be determined 
in magnitude by the link-polygon method (Art. 101). Some- 
times the link-polygon method is more troublesome than 
the following : — To find the push or pull in d e e'. We 
know (Art. 98) that the moment of the force in e e' about 
the point b is equal to the sum of the moments about b 
of all the external forces (for the forces in the directions 
a a' and c c' have no moment about b, since they pass 
through it). Let the algebraic sum of the moments of the 
external forces be actually calculated, multiplying numerically 
each force by its perpendicular distance from b. This sum, 
divided by the perpendicular distance from b to e e', will give 
the force in e e'. If the algebraic sum gives a moment tending 
to turn the structure about b against the direction of the 
hands of a watch, the force in e e' is a pulling force acting 
from E towards e', and therefore the piece d e is a tie. The 
engineer ought to practise this common-sense way of applying 
our fundamental principles. He will regret it if he fetters 
himself to graphical statics methods of working. 

It will be observed that if we wish to know the forces at 
any section of any loaded structure, we must consider that the 
parts of the structure on any one side of this section are in 
equilibrium. Thus, if A and b are the two parts of the structure, 
consider the equilibrium, say, of b. Now, b is kept in equili- 
brium by the external forces or loads which act on b, and by 



160 



APPLIED MECHANICS. 



the forces which act on b at the section. Of course, it is a 
which causes these forces to act on b through the section ; but 
in calculations concerning them we do not need to consider a 
or the loads on A. 

125. Loaded Links. — Let a c, c d, d e, and e b be four 
links hinged together at c, d, and e, and supported somehow by 

hinges at a and b, and, 
neglecting the weights of 
the links themselves, let x, 
y, and z be forces acting at 
the three joints, so as to 
make the links take the 
positions shown in Fig. 104. 
Take any point, o (Fig. 105), 
and draw lines o m, o n, o p, 
and q parallel to the links, 
and from any point, m, in 
m, draw m n parallel to 
the force z, np parallel to 
the force y, and p q parallel 
to the force x : then it is 
easy to prove that the lengths 
of the lines m n, np^ and f q are proportional to the forces z, y, 
and X, and the tensile forces in the links are proportional to the 
lengths of the lines o m, o 7i, o p, and o q. For it is evident that 
the three forces at e, keeping the joint in equilibrium as they 
do, must be proportional to the sides of the triangle o mn. If 
you put arrow-heads on om and on circuital with the one 
already on m n, you will see that the bars B E and D E do not 
push the joint e ; they 2^uU it and are tie-bars. Thus, then, 
the lengths of the lines in Fig 105 represent, to some scale, all 
the forces acting at the joints, c, d, and e. 

All so easy as it is to prove that the above proposition is 
correct, it is well to illustrate its truth in the laboratory. 
Let a, c, d, e, b be a string fastened to a vertical board at A 
and B, with loads x, y, and z applied at c and d by means of 
strings passing over pulleys. The string, instead of being 
fastened at a and b, may there pass over pulleys with balancing 
weights. Let the vertical board be covered with paper ; with 
a pin prick points on the paper showing the directions of the 
string everywhere, and transferring the paper to another board, 
find what is the pull in a c, c d, d e, and A B. We have no 




Pig. 105 



APPLIED MECHANICS. 



161 



actual test of the pulls in c d and d e, unless an ingenious 
student can introduce very light spring balances whose own 
weights are negligible. 

The student may compare this construction with exactly 
the same construction in Art. 97. He will see that in Fig. 104 
the resultant of the forces e b and z is in the direction d e ; the 
resultant of the forces e b, 2; and y is in the direction c d ; the 
resultant of the forces ^b, z, y and x is in the direction A c. 
Fig. 104 shows the 2^ositions, and Fig. 105 shows the amounts of 
these successive resultants, b e d c a is what we called a line 
of resistance. 

126. Loaded Chain. — If we want to find the pull in every 
part of one chain of a suspension bridge, and to draw the 
shape of the chain, it is first necessary to know the weight of the 
bridge at every place. This weight is probably supported by 
two chains, so, as we have only one chain to deal with, we only 
take half the weight of the bridge. We shall suppose that there 
is no long girder or other support for the bridge but the chain. 
It is usual to suspend the supporting beams of the roadway 
from the chain by vertical iron rods, placed at equal horizontal 
distances from one another. We may imagine the roadway to 
be as heavy at one place as another, so that the pulls in all the 
rods will be the same. Suppose there are ten rods, and in each 
a pull of 20 tons. Draw 
ten equidistant vertical 
lines (Fig. 106) to repre- 
sent the rods. We must 
get another condition ' 

before we can draw the 
chain. Let it be this, that the 
chain in the middle, where it is 
horizontal, shall be capable of with- 
standing a pull of 200 tons. Now 
draw H horizontally (Fig. 107), 
and make its length on any scale 
represent 200 tons. Make H A 
and H B on the same scale represent 
each 100 tons (if your chain is to 
be symmetrical), and divide them 
up, so that each portion represents 20 tons — that 
vertical load communicated to the chain by each 
Now join o with each point of division in A b. 




Fig. IOC. 



Fig. 107. 




is^ the 

tie-rod. 

Suppose 



162 APPLIED MECHANICS. 

that p (Fig. 106) is one point of support of the chain, draw 
p a (Fig. 106) parallel to o a (Fig. 107), a c parallel to o c, 
c d parallel to o d, and so on till you reach the point q, 
which I suppose to be on the same level as p. Of course, 
the points of support, p and Q, may be anywhere on the 
lines a P and m Q. It is quite evident from what you have 
already learnt that the pull in any part of the chain is 
represented by the length of the line from o, which is parallel 
to it in Fig. 107, and it is also evident that the chain will take 
this shape without any tendency to alter. Note that when all 
the loads on a chain are vertical, the horizontal coinponent of 
any of the sloping forces is the same as that of any other, being 
H. This is always called the horizontal pull of the chain. 

127. We began by assuming a pull of 200 tons in the part//i, 
where the chain is horizontal. We might have assumed a pull 
of 300 tons infh ; this would have caused the chain to hang in 
a flatter curve. Assuming a pull of 100 tons in/h, we should 
have obtained a greater difference of level between p and h. 

It will be found that in the present case, where the load is 
supposed to be uniformly distributed along the horizontal, the 
links would just circumscribe the curve called a parabola. 
With any other distribution of load they will fit some other 
curve than a parabola, but in any case you know now how to 
draw the shape of such a chain, and to determine the pull in 
any part of it. 

128. Arched Rib. — If instead of a hanging chain you wanted 
to use a thin arched rib to support your roadway, then if you have 
numerous vertical rods by which to hang your load to the rib, 
and if the distribution of the load is known, you can draw the 
curve of the rib in exactly the same way, but it will now be 
convex upwards, of course. With uniform horizontal distribu- 
tion of your load you will get a parabolic rib. The difference 
between the two cases is this : a slight inequality in your loads 
or a teuiporary alteration will only cause the chain to take a 
slightly different position for the time, and it will get back to 
its old shape when the old loading is returned to ; whereas the 
arch is in a state of unstable equilibrium, and as it is very 
thin, so that it cannot resist any bending, a slight change of 
loading will very materially alter its shape and it will get 
destroyed. Such a rib or series of struts is either stayed 
with numerous diagonal pieces or else it is made very 
massive, so that should the line like v h q (inverted)j which 



APPLIED MECHANICS. 



163 



is supposed to pass everywhere along its axis, deviate a little 
from this position, the rib may resist alteration of shape by 
refusing to bend. 

129. The load carried by an arch may either be hung from 
it by means of tie-rods, or else it may rest on the top of the 
arch, the weight being carried from the different parts by means 
of struts or pillars of iron, stone, or brick, or the arch may be 
levelled up to the roadway by means of a solid mass of masonry, 
or merely by one or two pillars of masonry, the roadway being 
carried on little arches from one to the other ; or there may be 
a filling-in of earth. It is rather difficult in a stone or brick 
bridge to say exactly what is the load on every portion of the 
arch, but it is guessed at, and a curve or line of resistance, such 
as p h Q, Fig. 106 (inverted), drawn. It is shown in Art. 368, 
that in a stone or brick arch it is dangerous to have the arch 
so thin that the line p A Q (inverted) passes anywhere outside 
the middle third of the arch ring. Thus, in Fig. 108 we have a 
section of a stone arch, the 
various stones or voussoirs, as 
they are called, being separ- 
ated by joints of mortar or 
cement. Now divide each 
joint into three equal parts 
and draw two polygons, m in m 
and n n n, marking out the 
middle third of every joint. 
Let us suppose we know the 
weight which each voussoir 
supports, including its own 
weight (it is usual to consider 
the arch as one foot deep at 
right angles to the paper), and 
let these weights be the 
weights w-^, w^, etc., shown 
in Fig. 108. I have taken a 
case in which these loads are 
symmetrical to right and left 
of the crown. Now draw the 
force polygon. Fig. 109 ; it 
vertical line, the forces beinj 




Fig. 109, 



happens to be all 
f all vertical. And 



m one 

now we 



come to the drawing of our line of resistance, but we are 
stopped at the outset by not knowing what is the thrust 



uu 



APPLIED MECHANICS. 



at the crown of the arch. The pull at the middle of oor 
susj)ension bridge chain was quite definite, but the thrust 
at the cro\s-u of the arch may be what we pleaise, and the 
arch will remain stable if the link polygon which we 
di*aw never passes outside the middle third of any of the 
joints.* Suppose we draw any symmetrical link jwlygon 
to begin with, by bisecting a k in u (Fig. 109), draw h o 
horizontal, and take o anywhere we please, o H will be the 
thrust in the crown of our arch, if this link polygon is the 
correct one. Join o a, o 1 2, o '2 3, etc. Start from any con- 
venient point in ic.^ Fig. 108, say e. within the space which 
contains the middle thirds of all the joints. Draw e d, Fig. 108, 



5 E .- 




F g. 110. 

pai-allel to o 4 5, Fig. 109 ; draw d c, c b, b a, a p, in succes- 
sion parallel to the corresponding lines in Fig. 109, and so also 
for E F, etc., to K Q. If any of the lines so dmwn passes out- 
side the space m m, u n, we must choase some other point E to 
begin at. and if we find that no choice of E ^vill allow the link 
|x)lygon to lie altogether within the space in Jii, n n, then we 
must choose another pole, o, in Fig. 109, until at length we find, 
as in the figure, a link ix)lygon, p E q, which cuts within th(» 
middle third of every joint. The lengths of the lines in 
Fig. 109 tell us the forces acting at the joints of Fig. lOS. 
Thus o a. Fig. 109, is the foi-ee p A, Fig. 108. the resistance 
of the abutment of the bridge. Again, the length of o -4 5 is 
the force acting in the direction e d between the stones e 
and D. 

130. Professor Fuller has made the work of dra\^-iug such a 
link iK)iygon very easy. In case the loads are all parallel to one 

* It is obvious also that tlie link i>olygon, wherever it crosses a joint, 
must make an angle so near a right angle with the joint that there can be no 
slipping or rupture by shearing there. 



APPLIED MECHANICS. 165 

another, it can be shown that if a number of link polygons are 
drawn in Fig. 108 for different lengths, o ii, Fig. 109, then the 
vertical distances between the points a, b, c, d, etc., are in the 
same proportion in all the link polygons."^ Beginning with the 
first load w^, draw (Fig. 110) A 1' 2' 3' 4' 5' e, the half of any 
link polygon corresponding to P A B c D E in Fig. 108. Divide the 
horizontal a s into any number of equal parts ; I choose six. 
Erect perpendiculars at A, 1, 2, 3, 4, 5, s. Draw horizontal 
lines AST and from 1 ', 2', 3', etc. ; draw any inclined straight 
line of convenient length, et; draw vertical lines from t, 1", 
2", 3", etc. From the points where the verticals a a', 1 1', 2 2', 
etc., cut mm and nn, draw horizontals to cut the correspond- 
ing verticals from t 1", 2", 3", etc. Join the points so found by 
the curves m m" and nn"; then, just as \h.3 straight line et 
represents a link polygon, m m" n n" represents the area bound- 
ing the middle third of all the joints, and any link polygon will 
be represented on the right hand side by a straight line. Now 
draw a straight line lying altogether within the space m m" n n". 
If you can draw several, then draw that one which is steeliest, 
in this case c t. Project this over to the left hand side, and 
you will find that you have the link polygon, which supposes 
the least thrust at the keystone. The corresponding force 
polygon has its o H less than the o H of A E in the proportion 
s c to s' E. The proof of this is easy. 

In the figure the possible line a! s' t' is so close to a s t that it is 
not shown. The proposition that the line of resistance must lie 
within the region of middle thirds will be dealt with in Art. 368. 
Fig. 109 shows the amounts of thrust between the voussoirs, and it 
is worth while comparing these forces with the areas of the joints. 
In Art. 368 the strength of such joints is more particularly studied 
It is also worth while to consider whether the line of resistance 
may not be made to pass outside the region of middle thirds by a 
possible unsymmetrical load. 

But we cannot give a proof of the proposition on which the 
whole work dejiends. If any line of resistance may be drawn to 
pass inside the region of middle thirds, the arch is stable, and the 
steepent of such lines is the actual line. We have a lecture model 
in which a number of rounded blocks of wood cdefghij 
represent voussoirs, the fixed parts a and p. being abutments. We 
can l(jad these voussoirs in all sorts of ways. When the loading is 
changed, the voussoirs will be seen to roll on one another into new 

* See Art. 349. Proof. Each link polygon is really a diagram of bending 
moment, supposing the loads were acting on a horizontal beam, and the scale 
of each diagram is proportional to the length of oh. 



166 



APPLIED MECHANICS. 



positions ; and if the points of contact are joined, we have the line 
of resistance. 

^Tien the loading is symmetrical, loads on the haunche? cause 
the line to rise at the haunches and get lower at the crown, and a 
load on the crown reverses this effect. A student may see on this 
model how change of loading causes such a yielding in the arch as 
produces pressures at the abutments just suited to equilibrium, 
pro^dded only that a line of resistance cuts all the joints. But we 
must confess that the sudden step in the reasoning from this to the 
mere statement of the above proposition does not satisfy us. At 
the actual plane joints of an arch, changes occur when the load is 
changed, but these changes are very different from the changes 
that occur in the model ; and it seems to us that the only 
legitimate method of study is that of assuming that a masonry 
arch behaves exactly like an arched rib of iron fixed at the ends, and 




Fiar. l\\ 



this is studied in Art. 380. The line of resistance being foimd as 
for an iron arch, the masonry arch must be so designed that this 
line is kept within the middle third. 

In a few cases in Germany an attempt has been made to build 
masonry arches hinged at the ends, and, indeed, also with a hinge 
at the crown. This can be done by bedding the masonry at these 
places upon iron. A quasi hinge has also been employed by 
introducing plates of lead at the two joints at the springings, and 
one at the crown, the lead only extending over the middle thirds of 
the joints. 

131. Buttresses. — To find the force which acts from one 
stone to another in a buttress, it is necessary to know the force 
acting on every stone from the outside, and also tlie weiglit of 
the stone. Find the resultant of these two forces for each 
stone, and draw the link 'polygon whose first side is the force 
on the top stone. In Fig, 112 fabkp is the link polygon so 



APPLIED MECHANICS 



167 



drawn. Each side of it shows the resultant of the forces 
acting at every joint, and the length of the corresponding line 
in Fig. 113 shows its amount. Thus the resultant of the 
forces acting at the joint s t is shown by the position of the 
line B K, and its amount is shown by the length of the line o t 
in Fig. 113. 

If we see that any of the sides of the link polygon passes 
outside the middle third of the corresponding joint between 
two stones, we know that part of that joint will be subjected 













/F 








y 


m 


1 .4 

1 ,' 7^ 


V 










',' 


;-'a 








z 




4 


■-4 

n 




w 






1 ■■' 








s 


7?t 


Ir 


n^ 




_T 






< 


TV 








Fig. 112. 



Fig. 113. 



to tension, a condition to which we suppose that a common 
masonry joint ought not to be subjected.* 

The student ought to work examples in which, besides the 
force F acting on the top stone, there are forces acting on the 
other stones, due, say, to the pressure of water (Art. 173), or 
to the pressure of earth (Art. 293). 

132. To return to the hanging chain. If the total weight of a 
chain and all the vertical loads upon it is w, and if its ends are sup- 
ported at two points, a and b, and at these points it makes angles 
o and ^ with the horizontal, the tensions there are represented by 
the two sides of a triangle, which are parallel to the directions 
there, the thu-d side, representing w, being a vertical line. Let 
the student sketch examples, 

* In many cases it will be found well to magnify all the horizontal 
components of all the forces, magnifying the horizontal dimensions of all the 
stones in the same proportion. In this way the points in which each side of 
the link polygon cuts each joint may be found more accurately. 



168 APPLIED MECHANICS. 

The tension t at a is evidently w cos. /8/sin. (a + B), and the 
tension at B is W cos. a/sin. (a + B). 

If at B the chain is horizontal, then t at a is w/sin. a, and the 
tension at b is w cot. o. Call this t^. 

If X is horizontal distance of any point in. the chain from b 
(where the chain is horizontal) and 7/ is vertical height above b ; if 
the load on the chain between b and any point a is ivx where w is 
the load per unit length of horizontal projection of the chain [here 
I assume that the load is uniformly spread ho7'izontalIy, and this is 
nearly the case in any very flat chain or telegraph wire], then, as 
before, if a is the angle of inclination of the chain at a, we have 

Tq = tvx cot. a ; or, in the language of the calculus, t^ -- = wx ; 

and it follows that the shape of the chain is 3/ = |- — a?^ .... (1), 

To 

a parabola with vertical axis whose vertex is at b, the lowest point. 
Given the heights of the two points of support of such a chain 
above its lowest point, and also the horizontal distance between 
them and iv, it is easy to calculate the value of Tq and everything 
else. 

EXERCISE. 

In a suspension bridge of 800 feet span and 80 feet dip, the weight of 
the i^latform, chains, and rods, etc., is 2|- tons per foot run, what will be 
the horizontal tension in each of the two chains, and the tension in each 
at the point of support at the piers ? Ans., 1125 tons ; 1212 tons. 

133. In a telegraph wire of span I, the points of support being at 
the same level, if the dip (supposed to be small compared with I) is a, 
the weight of wire being w^ or tvl nearly, it is easy to show that the 
tension, which I shall here call p, is wP/8a or ^\l/8a at the lowest 
point, and the tension elsewhere is not much greater. Now sup- 
pose s, the whole length of a wire, to be b {I + k0 + 13 p) where 
b is its length at the lowest temperature, and this we shall call 
the zero of temperature, under no tension, 6 being the temperature 
Centigrade at any other time above the zero, k the coefficient of 
expansion, and j3 a constant easily determined when one knows 
Young's modulus for the material and the cross-section of the wire, 
just as w is known from the material and cross-section. If a is the 
dip, find the tension p, and find the length of the wire. The 
length of a parabolic curve is easily found by integTation, but 
X -\- 2 1/1^ X gives the length from the origin to any point with 
sufficient accuracy for such purposes as the present, [Check 
this statement when you have leisure.] So that length of wire is 
I + 8 «73 I. 

If ttf^ is the dip at the zero of temperature we are led to the 
result 

" = <"»-'')|§^ + |(- + "»'|----« 

where a = ajl, Oq = a^fl. 

If, now, the span I and w (the weight per unit length of a wire) 
are given, and if p^ be taken as the greatest pull to which the wire 



At>I>LlED MfiCflANlOS. l6& 

ought to be subjected, this will be at the lowest temperature, 
»9-hich we here call zero. Then Oq = wlj^ Pq. 

Choose now some gTcater yalue of a, and from this, by (1), 
calculate 0, and note also that p = wlji a ; so that it is easy to 
make out a table showing- a and p for various temperatures. 

This calculation may be regarded as a mere exercise for 
students, and yet there may be occasion for its use in practice. 

The linesman who puts uj) telegraph wire pays very little 
attention to the result of such a calculation. He gives such a dip 
to his wire as seems good to him, making it somewhat less in 
winter than he would in summer. However much he may have 
killed the wire already, it will permanently stretch a little more if 
a very cold night comes, without real hurt to its material. 

If a chain is uniform, and is loaded only with its own weight, 
and if we desire an exact answer instead of the above approximate 
answer for fiat chains or wires, we let ^ = w s where w is the 
weight of unit length of the chain and s is the length of chain 
, from B, its lowest point, to any point a. Then 1^=^ ws cot. o 

leads, by the calciilus, to the result y = -I e - + e '^\ where 

A \ c c / 

t(; c = To, and c is the vertical height of b above the origin. This 

curve is called the catenary. 

As we have t = t,, tan. a, it is easy to find t. It is also easy 

to find s in terms of x and y. The properties of this curve give 

very easy exercises in the calculus, and so they are well known. 

FIDDLE STRING-. — Neglect the weight of a string whose mass is m 
per unit length. When ^dbrating in a plane at any place at the distance x 
from one end let the displacement be y from the equilibrium position. 
At the ends of a short length dl, the resultant force tending to bring 
the mass in U back to the equilibrium position is t. ZQ if 10 is the angle 
between the tangents at the two ends, or t. Zl x curvature. The string 
being everywhere very nearly straight, we may take the curvature to bo 
d^y 



d x^ 



' (See Art. 25.) This force, divided by the mass m SI, is acceleration. 



or - j-^; ^^<i hence 

'- di ^ ^dl = «• 

I was once at a great loss to know how to measure the tension in each 
span of a tight telpher line ; rods of round |inch steel, each 200 feet long. 
At length I saw that I could use the fiddle-string principle and, if the rod 
is set vibrating in the simplest manner in a vertical plane, the time of _a 

complete (up and down) oscillation is very nearly, t seconds = 21 .a/ -. 

This principle, borrowed from acoustics by engineering, enabled labourers 
to test the line, sj)an by span, every day. If there were less than 17, or 
more than 19 complete oscillations in a quarter minute, the span needed 
to be tightened or slackened. 



^ 



170 

CHAPTER TX. 

HYDRAULIC MACHINES. 

134. Hydraulics. — Hydraulic machines are very wonderful 
to people who observe their action for the first time. Ancient 
drawings show armies of slaves dragging on ropes to lift a single 
weight. Three hundred years ago, Fontana raised an obelisk at 
Rome with 40 capstans, worked by 960 men and 75 horses. A 
few years ago a sioiilar obelisk was raised in London by four 
little 100-ton hydraulic jacks, each of which can be worked by one 
man. Large modern guns would be almost impossible to work 
with any other machinery. If you go to any large docks you 
will see how, by the manipulation of a few handles, a boy can 
remove heavy objects rapidly from ships and place them on the 
dock by means of an hydraulic crane. Visit any large steel 
works, and you will see great armour-plates and Bessemer 
converters and their appliances passed about nearly as readily 
as small objects are moved by blacksmiths and moulders. The 
steam-hammer, powerful as it is, is giving place to hydraulic 
forg-ing and squeezing machinery, because the new forces are 
enormously greater, and their effects can be more uniformly 
distributed over large masses of metal, leaving them more 
homogeneous. Visit the Victoria Docks, and you will see 
vessels of 3,000 tons raised out of the water on a floating grid- 
iron and towed off for repairs. Visit the River Weaver, in 
Cheshire, and you will see sections of a canal rising and falling 
50 feet with canal boats, which are no longer delayed for hours 
in floating through a flight of locks. Instead of bringing great 
iron girders near a riveting-machine, we now take little rivethig 
machines to the girder, and Avork them in any position through 
small flexible pipes from a distant steam-engine and pumps. 

135. Hydraulic Press. — An hydraulic press is a machine 
which enables large weights to be lifted or great pressures 
exerted, but in which, instead of levers and wheels, we use water 
to transmit the energy. In Fig. 114 we have rams of large 
and small cross sections a and a square inches, and weights 
(neglecting their own weights) r and e. Because the vessel c is 
full of water which cannot escape past the rams on account of 
two leather collars or other packing at d and g, the velocity 
ratio is A/a. Thus if A/a is 100, and if a falls through 100 



APPLIED MECHANICS. 



171 



inches it displaces ax 100 cubic inches of water. We assume 
that the water cannot escape, and that all the room needed is 
procured by the lifting of a; then a must lift one inch to make 
the necessary room. 

In actual specimens, instead of 
one large vessel c, we usually have 
vessels round a and a slightly larger 
than themselves, called a press and 
pump-barrel with a pipe connecting 
them. The friction at a leather collar 
is sometimes as little as one per cent, 
of the weight on the ram, and some- 
times as much as 5 per cent. Notice 
that the amount of the axial force R 






which may be exerted, neglecting 

friction and weights of water, does not Fig. 114. 

in any way depend upon the shape of 

the end of its ram, or where it is, or the direction of the 

ram's motion. 

136. The internal construction of an hydraulic press is shown 
in Fig. 115. Three men press, each with a force, say, of 60 lbs., 
on the end of the lever G, whose mechanical advantage is, say 
20, and hence the plunger, Ea, is pressed downwards with a 
force of 3,600 lbs. Just consider for an instant what is the 
condition of things before E moves. There is a ram, r, which 
carries a heavy weight, the weight to be lifted. Observe that 
this ram is wanting to fall, but it can only fall into the vessel d. 
Now the space between the vessel, or press, and the ram, and 
all the space in the tubes, T, is filled with water which has no 
means of getting away. It might get away by the little valve 
p, but that valve can only open upwards, and the more the 
water tries to escape, the more it really closes the valve, just as 
a closely-packed crowd in a panic keeps an inward-opening 
door closed, by which they might otherwise escape from a 
theatre. There is no escape for the water on the pump side ; 
there is just as little on the other side, for you see that the 
water, if it escapes into the the space N, finds that it has still 
to get past the leather collar, which is, however, so placed and 
shaped that the greater the water pressure the tighter the 
leather fits the ram. Fig. 117 shows such a leather collar. 
There is then no escape for the water, and when this is the case, 
no matter what weight is placed on the top of the ram, it 



172 



APPLIED MECHAXICS. 



cannot fall. The falling of the ram would mean some escape 
of the water ; but as there is no e5ca})e for the water, the ram 
will fall no more than if it were suppoi"ted on some quite rigid 
material. 

137. I have been supposing that a certain quantity of water 
will absolutely refuse to occupy a smaller space, but this is not 
quite correct. We know that if it were air that fiUed the space x. 




instead of water, and there were no escape for it, the ram would 
fall when a greater weight was placed on it ; for although the 
air cannot escape, it is contented to occupy a smaller bulk. 
I have been supposing that this does not occur in water, and 
that water will refuse to go into a smaller space, whatever the 
pressure. This was an old notion which people deduced from 
the famous Florentine experiment. A hollow globe of gold 
was quite filled with water, and was hennetically sealed. It 
was then beaten to diminish its cubic contents, and the result 
was that drops of water made their appearance on the surface, 
having oozed out through the pores in the gold rather than 
submit to the lessening of the total bulk. 



APPLIED MECHANICS. 173 

I am told that a lecturer at Chatliam subjected the water 
inside a cast-iron shell to so much pressure that it came through 
the pores of the iron, and appeared as a fine spray or mist on 
the outside, and soon afterwards the shell burst, or, rather, fell 
gently in pieces. But with a piezometer it is easy to show 
that water and all other substances will submit to a diminu- 
tion of their bulk when subjected to a pressure ; and we find 
that this diminution for water is l-20,000th of its total bulk 
for a change of pressure of one atmosphere, or 1-7 0th of its 
volume for a pressure of 2 tons per square inch, such as we 
find in hydraulic presses — that is, 70 cubic inches become 69. 
Now this diminution in bulk is far too insignificant to be 
of much practical importance in hydraulic machines, and 
we may take it for granted that whatever weight we place 
on the ram, it will not perceptibly fall, because the water 
refuses to become smaller in bulk, and because it cannot 
escape anywhere. 

We understand that it fries to get away j it is trying to 
burst the thick cast-iron press ; it is trying to burst the pipes 
and the pump. Before it will burst the pump, it will open the 
safety valve h, and escape ; but we can assume for the present 
tliat the pressure never reaches the bursting pressure of the 
arrangement. 

Now the labourer acts on the lever, forcing down the 
plunger E a. The water in the pump-barrel is j ust like the water 
in the press ; it tries to escape, it tries to burst the pump-barrel, 
it resists the motion of the plunger. It tries to escape through 
the valves f and h, and it will open the valve f, and pass 
through, if the labourer acts sufficiently on his lever ; but if the 
water passes through F, the ram A R must rise, however great 
the weight may be that is pressing it down. The question is, 
then, what force on the plunger e (X is sufficiently great to cause 
motion — that is, to cause the water to pass through the valve 
F, and so make the ram a r rise 1 

Suppose that the plunger e a is, one square inch in section, 
and that one inch more of its length is forced into the pump; 
evidently the metal takes up the place of an equal bulk of water, 
or one cubic inch. This cubic inch of water has found one 
cubic inch of room for itself somewhere else. As we suppose 
no greater compression of the water, and no yielding of the 
sides of the press, it is evident that one cubic inch of the ram 
must leave the press to give the water the space it must have. 



174 APPLIED MECHANICS. 

Now, if the ram is 100 square inclies in section, then 1-lOOth 
of an inch of its length contains one cubic inch in volume. If 
the ram lifts through the distance 1-1 00th of an inch, it will 
therefore leave one cubic inch of room behind it. 

We are not concerned with the shapes of the ends of the 
plunger and ram ; we know that if one more inch of the 
plunger enters the water, 1 -100th of an inch of the ram must 
leave the water; that is, the relative speeds of plunger and 
ram are as one inch to 1-1 00th of an inch, or as 100 to 1. The 
plunger must move 100 times as quickly as the ram, and by the 
law of work, if there was no friction, a force of one pound on 
the plunger would balance a force of 100 pounds on the ram. 
We know the mechanical advantage, then, if there were no 
friction, in the portion of this machine from plunger to ram, if 
we know how many times greater is the area of the ram than 
the area of the plunger. 

138. Different experimenters give diflferent results as to the 
loss of energy, and therefore of mechanical advantage, in friction. 
Rankine states that there is 20 per cent, of loss. Mr. Hick 
found results less than one-tenth of this. It is kno\vn that in 
accumulators, the pressure which is great enough to lift a 
load, being, say 1,010 lbs. per square inch, if the pressure is 
reduced to 990, the same load will fall; thus only one per cent, 
of the energy seems to be wasted in friction at the leather 
collar, when the motion is slow. It is probable that there is 
always less than 20 per cent, of loss of energy altogether in harge 
hand-worked presses. As an instance of this greater efficiency 
of hydraulic machines, a platform weighing 12 tons had to be 
lifted, and a little hydraulic jack was placed under one corner, 
a screw jack under the other. One man was told off to work 
the hydraulic jack and thi'ee men to the screw jack. The man 
at the hydraulic jack, with one hand in his pocket, would pimip 
a few^ strokes and then quietly wait a little, whereas the three 
men were hard at work all the time. 

My students in Japan used to employ a little hydraulic 
press to crush bricks, stones, and wood, and it w^as roughly 
assumed that the friction at the glands was insignificant. Of 
course, we only made this assumption when we wanted rapidly 
to get a rough idea of the relative strengths of materials to 
resist crushing. Still, it was a sort of thing that we should 
not have been able to do with any other except a lever 
machine, and I am inclined to think that there is more 



APPLIED MECHANICS. 



175 



inaccuracy with many lever-testing machines than there was 
with my hydraulic press. 

Example. — A three-ton hydraulic jack, like Fig. 116, not in 
very good order, was examined in my laboratory ; weights A 
were hung at the end of a lever, and we calculated r the 
equivalent weight which the jack l was actually supporting. 
The handle was replaced by a wooden sector, to which the 
weight B gave a slow motion, after one or two strokes had 
been made by. the experimenter to get things into a steady 
condition. The weight b, which would slowly overcome a 
given R being noted, the experiment was repeated many 




Fig. 116. 

times with other loads, and the values of R and b tabulated 
and plotted on squared paper. We found the result : 
B = -024 R +2-2 where b is the effort, in pounds, applied 
at the end of the handle, and r, in pounds, is the load -on 
the jack, not including the weight of the ram itself. The 
mechanical advantage of the lever was 14-75 ; the ram was 
2 inches in diameter and the pump plunger 1 inch, so that the 
total velocity ratio was 14*75 x 4, or 59. If there had been 
no friction the effort, b^, would have been r -r 59 ; hence the 
efficiency is — 

^=4'-=^^-C024r + 2.2) = 



59 



1-42 R + 130 



1-42 + 130 



Thus when r is 3 tons, or 6,720 lbs., the efficiency is 0-69. 
There can be no doubt that the loss of energy mainly occurs by 



176 



APPLIED MECHANICS. 



solid friction, and so we are led to ask, where does friction 
occur in the hand- worked hydraulic press 1 It occurs at 

1. The rubbing surface at the fulcrum of the lever. This 

is the fiiction of solids. 

2. The rubbing surfaces of the two glands; one, that of the 

plunger, the other, that of the ram. Here, again, we 
have friction, as if between solids. 

3. Everywhere in the water where there is motion. This 

is fluid friction, which is quite different from that of 
solids. 

139. Figs. 117, 117a, show sections of the leather collar 
used in presses. It is made from the best leather, softened in hot 





Fig. 117a. 



Fig. 117. 

water, and pressed between cast-iron moulds to its present 
shape, the pressure gradually increasing. It is then left for 
several days under pressure in the mould. When it is in its 
place in the press the water gets behind the leather, and presses 
it tightly against the ram. The friction seems mainly to occur 
at the part A, and increases as the jjressure of the water 
increases. There does not seem to be much friction at the 
portion between A and B, and the efficiency of the press is 
but little altered by making a b greater or less. It is, 
however, asserted by some makers of presses that the dis- 
tance A B is of importance ; but I rather think that this is 
on account of deterioration. The part b a is constantly being 
in states of tension and compression, and is liable to crack. 
When the leather deteriorates, much time is wasted in 
renewing it. 



APPLIED MECHANICS. 177 

140. Hemp-packing is invariably used instead of the leather 
collar at low pressures ; some manufacturers never use hemp 
or cotton when the water has a greater pressure than 700 lbs. 
per square inch, but others use hemp to 2,000 lbs. per square 
inch. 

Our subject is too large to allow me to enter into many- 
details as to the advantages and disadvantages of leather, 
india-rubber, and gutta-percha for packing purposes. There is 
a great divergence of opinion, and I believe that much of the 
evidence against one or another material is based on the bad 
preparation of the materials against which the evidence is 
given. 

141. In Fig. 117b you will see the form of india-rubber cup 
used in hydraulic plungers, jacks, and bears. It is moulded in 
the form shown, and is fastened to the end of the ram by means 
of a screwed bolt and metal washer, as shown in the figures. 
The water pressure keeps it tight against the cylinder at the 
edge, and this is where the friction occurs. 

The quasi-solid friction between lubricated leather or hemp 
or rubber and metal, follows the laws of solid friction, and 
we have now described what friction of this kind occurs in 
hydraulic presses. The remaining source of loss of energy is 
the fluid friction. 

The flow of water even in the narrow passages past the 
valves may be very slow, and therefore the fluid friction may 
be as small as we please. So small, indeed, is the fluid velocity 
that when oil is used the loss of energy is much the same. 
Probably honey or tar would not give very ditFerent results ; 
but the more viscous mixtures of tar and pitch would be 
unsuitable, because in these the fluid friction, at even such 
small velocities as we have to consider, would be considerable. 
Even solid pitch would, however, act as a fluid, and might give 
the same mechanical advantage as water if we made only one 
stroke of the plunger in a month. 

142. In applying the law of work to our machines, equating 
the energy given to, and the energy given out by, each machine, 
we assume no store of energy in the machine. In the press, even 
if we disregard the compression of the water and the elastic 
yielding of the press, we must remember that the ram itself is 
lifted, and also that some water is lifted in level. The lifting 
of the ram is easy to take into account ; consider it part of the 
weight to be lifted. In oi-dinary presses it is of no great 



178 



APPLIED MECHANICS. 



consequence. But when the dead weight of the ram and other 
things is greater than the weight lifted usefully, as in ware- 
house and hotel hoists, it cannot be neglected, and it is usually 
easily balanced. In very high hoists the weight of water 

which changes its level 
becomes important. (See 
Art. 169.) 

143. Fig. 118 shows a 
section of a lifting jack 
with its ram r lifting any 
weight which may rest on 
J or I. Notice how the 
ram r fits the press m, and 
is made water-tight by the 
india-rubber dish L. A 
handle or lever is attached 
at H, and when worked 
causes great pressure on 
a projection or cam K, 
which communicates with 
the plunger v^^ of the 
pump, and gives it a down- 
ward motion, pressing the 
water in the pump-cham- 
ber through the valve Y^ 
and the passage in R to 
the press m, which rests 
on the ground. A small 
pressure on the lever thus 
forces more and more 
water into the press, and 
necessitates the upward 
movement of the ram R. 
The upward motion of the 
lever causes a partial 
Fig. 118. vacuum in the pump- 

chamber as the plunger 
V| is withdrawn, and the pressure of the air and liquid in 
the cistern c overcomes the resistance of the spring on the 
inlet valve v^, which is really a part of the pump plunger, 
and thus effects a passage for the water into pump-chamber. 
During the downward stroke the previously-described operation 




APPLIED MECHANICS. 



179 



is repeated. If we want to lower the weight we open 
the lowering screw s, and allow the water to return from the 
press M to the cistern c, the ram falling. 

144. The punching-bear, shown in section in Fig. 119, is 
similar in construction. Theplunger, pump, valves, etc., are much 
the same as in the last case. The upper part of the stout ram f 
terminates in an india- 
rubber dish, which is 
fastened by the washer 
and a bolt passing 
through the middle 
of the washer into the 
ram. 

As we work the 
top lever a, the ram 
F holding the punch 
is pressed down by 
water through a valve 
arrangement exactly 
similar to that in the 
lifting jack. 

145. We are now 
well aware of the way 
in which the water in 
these machines acts. 
It is nearly incom- 
pressible, and tries to 
find an outlet in every 
direction. Consider 
what occurs from par- 
ticle to particle of the Fig. ii9. 
water. Each particle 

presses on all its neighbours, because they all press in upon 
it, and it presses equally in every direction. 

Wherever the water comes in contact with a solid surface, 
it presses normally against the surface. There can be no such 
thing as oblique pressure in water when the water is at rest, for 
oblique pressure means a force partly along the surface, and 
this would imply some frictional resistance to sliding, which we 
know cannot occur in water at rest. It is evident from our 
discussion of the hydraulic press as a machine in which there 
is no store of energy, that on every square inch of the solid 




180 



A.PPLIED MECHANICS. 



surface touchtcl by the water there must be the same foi'ce 
acting, the water tending to escape everywhere ; and when we 
consider the whole case mathematically, we hnd that every little 
interface separating any two portions of water is acted on by 
this same pressure per square inch 

146. We shall be perfectly safe in all our notions of fluid 
pressure if we consider each particle of water to be a very small 
being, greased all over, so that it cannot possibly resist sliding 
past its neighbours. It can press normally against a wall or 
against any surface, but there cannot possibly be a tangential 
or frictional pressure between it and a wall, 
because it is well greased. All the water is 
trying to escape, and the total pi'essure on 
any surface is evidently proportional to the 
number of water particles pressing against the 
surface. Hence, if we have a piston A and 
a piston B (Fig. 120), the total pressures on a 
and B are simply proportional to the areas of 
the cylindric tubes in which they can move. 
Evidently it is of no importance whether the 
inner surface of a piston has projections or not. 
Thus there is the same total vertical pressure on piston m and 
on piston n. Fig. 121, if the cross-sectional areas of their 
cylinders are the same. 

Everything depends on this: will they leave the same empty 
space behind them if each of them moves one inch] — and we 




Fig. 120. 











know that in each case the empty space is simply the volume 
of one inch of length of the cylinder. 

147. In the same way, although the end of the ram may 
be curved, as in Fig. 122, and is therefore being acted on by a 
series of pressures normally to the curved surface, as in the 
figure, it is easy to show that the resultant action of the.se is in 
the direction A B, and is really the same as if the ram had a 



APPLIED MECHANICS. 



181 




Fig. 122. 



Eat end. Every one of these forces has a horizontal tendency, 
more or less, and when we leave out of account these horizontal 
actions, we get the same vertical result for all 
shapes of ends. 

You will understand this better, perhaps, 
if we consider a vessel, a, b, c (Fig. 123), to 
be filled with fluid at such a great pressure 
that we can neglect the pressure due to the 
fluid's own weight. We are assuming now 
that we can do this in the hydraulic press and 
in steam boilers. The pressure on the surface 
everywhere is shown by the arrows. 

It is evident that the total horizontal force on the curved 
surface A c b is exactly equal and opposite to the total hori- 
zontal force on the flat surface A b, because if there were on 
the whole more force on one than the other, the 
vessel would move bodily, an idea which is absurd. 
Hence, when we want to find the total horizontal 
force on the curved surface, we never dream of 
going into the long calculation which you might 
think necessary, for it is simply equal to the area 
in square inches of the Jlat surface A B, tnultiiolied 
hy the pressure per square inch. 

148. Suppose we want to find the horizontal burst- 
ing tendency of the egg-ended boiler m n, Fig. 124, 
that is, say, the force tending to burst it by direct pull of the 
iron at the section A b — we do not trouble ourselves with the 
shape of the boiler anywhere ex- 
cept at A B itself. The bursting force ^ 
is the inside area of A B in square ^ 
inches, multiplied into the pressure 
per square inch. The area of the iron 
in the section a b is exposed to this 
pull. These are the two important 
facts to be remembered. Consider 
any section whatsoever of a boiler, 
or ram, or pipe. Eemember that 

the fluid pressure is calculated over the whole area. The 
resistance of the iron is only calculated over the actual 
sectional area of the metal. 

Thus, if we want to find the tendency to burst along such 
a section as a b, we take the total inside area of the section 




Fig. 123. 



I 
124. 



182 APPLIED MECHANICS. 

multiplied by the pressure, and this is equal to the stress 
in the iron all along this section, multiplied by the whole 
sectional area of the iron. 

In a cylindric boiler or press, which is everywhere of the 
same thickness, it is easy to show that if we neglect the eflfect 
of the ends, the tendency to burst laterally is tioice as great 
as the tendency to burst endwise. 

Thus in the endwise bursting, if p is the bursting pres- 
sure in pounds per square inch, /the tensile strength ol" the 
material in pounds per square inch, r the radius of the boiler, 
and t the thickness of metal, the total force tending to produce 
bursting is the area of the circular cross section, 3-14 r^, multi- 
plied by p, and the total force resisting fracture is the circum- 
ference of the circular cross section, 6*28 r, multiplied by t and 
by/ Hence — 

3-14 r2j9 = 6-28 rft, 

or the bursting pressure p := '2ft-i-r. 

Again, if a boiler is I inches long, the total force tending to 
burst the boiler laterally is the area I times the diameter of the 
boiler, multiplied by jo, or 2 r ^ p, and the total force resisting 
fracture, if we neglect the ends, is the area of the iron 2 I t^ 
multiplied by/ Hence — 

2rlp = 2 Itj, 

or the bursting pressure p =ft -r r. 

Hence it would take twice as much pressure to burst the 
boiler if we assumed it to burst endwise. We always calculate 
the strength of a pipe or boiler on the second assumption 
therefore, and we have the rule : — The bursting pressure in 
pounds per square inch is equal to the tensile strength of the 
metal in pounds per square inch, multiplied by the thickness 
of the metal in inches, divided by the radius of the boiler or 
pipe in inches. 

149. When the press is thick, as we find it in an hydraulic 
machine, it is rather more difficult to calculate the bursting 
pressure, because the tensile strain is not distributed uniformly 
over the section at which there is a tendency for rupture to 
occur. The inside portions of the metal near the water are 
subjected to more pulling forces than the outer portions. 

The mathematical part of this subject will come before us 
in Art. 275. The subject is important and needs to be intro- 



APPLIED MECHANICS. 183 

duced here also. The result of the mathematical investigation 
is this : — Neglecting the strength of the ends, the bursting 
pressure, multiplied by the sum of the areas of the outer and 
inner circles of the cross section of the cylinder, is equal to the 
tensile stress which the metal will stand multiplied by the area 
of the metal exposed in such a cross section. This rule you 
will find applicable to thin cylinders as well ; it is the same 
rule as the one already given for thin cylinders. You see that 
if the fluid pressure is equal to the tensile strength of the metal, 
no thickness of the cylinder can prevent its bursting. 

The investigation might perhaps have some use if iron 
when it leaves the foundry had all through its thickness the 
same qualities, and a perfect absence from strain. It is good to 
remember that an iron casting when it leaves the foundry, 
although not quite so curiously strained as a Prince Rupert's 
drop or toughened glass, is yet in a state of strain which we 
know very little about. 

The above investigation shows that in a thick cylinder there 
is an exceedingly great difference in the tensions of the inner and 
outer portions of the metal of a gun or hydraulic press. Eor 
the purpose of producing a more uniform state of stress, the 
inside portions of the metal are often chilled ; that is, the metal 
in the inside is very quickly cooled after it is cast. The result 
is that these portions are virtually in a state of great compres- 
sion, and hence, when the regular strain occurs through water 
pressure, there is a considerable tensile yielding in the inner 
portions of metal before the bursting tensile stress is reached 
there ; and hence, with a proper amount of chilling, it is 
possible to have the tensile stress in the metal nearly uniform 
when fracture is ready to occur. In such a case as this, the 
calculation of the bursting pressure of a press is just the same 
as if it were thin. We shall return to this subject in Art. 275. 

It will sometimes surprise a student to see how much 
pressure is withstood by a wooden pipe wound round its outside 
with hoop-iron, or a hollow cylindric masonry or concrete 
reservoir with strong tightened chains or rings of iron on its 
outside. [The engineer need not fear deterioration of iron 
imbedded in cement. My brother, who has great experience, 
tells me that burying even one end of a bar of iron in cement 
has the power of preserving all of it from ordinary weathering 
or galvanic action.] 

150. Cast steel is now getting common for the cylinders of 



184 



APPLIED MECHANICS. 



small jacks and bears. The large presses used in warehouses 
are, however, still made of loam-moulded cast iron, as all presses 
used to be in my apprenticeship days. But in a great deal of 
large press-work cast steel has come largely into fashion, its 
tensile strength being so much greater than the strength of cast 
iron as to make a most extraordinary difference in the outside 




Pig. 126. 

sizes of the cylinders. Thus, for example, m a certain cotton - 
press, the use of steel enabled three cylinders to be placed side 
by side in a space which would only have allowed one to be 
used if it had been made of cast iron. 

Fig. 125 is a drawing of an hydraulic press, which gives a 
fairly good idea of the press used in warehouses for bales of 
linen and Manchester goods, linen-yarn, etc. The ram is usually 
about 10 inches in diameter, and the working pressure of the 



APPLIED MECHANICS. 185 

water from 2 to 3 tons per square inch. As the area of the 
cross section of the ram is 78 square inches, this means a total 
pressure of upwards of 230 tons. When used for packing hay, 
for expressing oil from seeds, and for other purposes, alterations 
are made in the shape, length of standards, and, indeed, in the 
size of all the parts. In expressing oil, for example, there 
would be an alteration in the arrangement of the table or 
platten and the head. Sometimes the table becomes a piston, 
fitting into a cylinder attached to the head. Usually, however, 
the arrangement is what you see. If yarn has to be pressed, it 
is placed in a great oak box, bound with iron, running on 
wheels. It is run in between the columns. The pumps work, 
a movable bottom of the box rises up, the pressure on the yarn 
gets greater and greater, and when about 230 tons is the total 
force, the yarn, very much diminished in bulk, is tied up, the 
ram descends, the great box is rolled out, and another loosely- 
filled one is rolled behind it to undergo the same process. 

For warehouse use the pumps are often worked by hand, 
but even thirty years ago I remember that orders were gene- 
rally for pumps worked by power ; that is, driven from shafting 
and worked by cranks. In this case it was common to attach 
a series of these presses to one set of pumps, all the presses 
in a w^arehonse being fed from a long pipe, each having its 
separate valve. 

151. You will see that, although in hydraulic jacks for lifting 
heavy weights the weiglit on the ram is the same during the 
whole of any operation, this is not the case in baling presses. 
Thus, for example, in the Indian cotton trade, governed as it is 
by the Suez Canal regulations, it is necessary to press cotton so 
compactly that it is like a piece of oak, and it can be planed up 
like oak. 

During the early part of the operation the pressure on the 
ram is therefore small, becoming very great towards the end ; 
and it is the greatest piessure, of course, which decides the 
relative sizes of plunger and ram. If the ram were made to 
rise quickly at the beginning, and more slowly towards the end, 
it is obvious that there would not only be a saving of time, but 
a more regular doing of work. 

In hand presses it is usual to change the fulcrum of the 
lever, so that a labourer may work more rapidly at the begin- 
ning. It is also common to use a large pump plunger in the early 
part of a pressing operation, changing it to a small one at the 



186 APPLIED MECHANICS. 

end ; or to use two of equal size, throwing one of them out of 
gear towards the end of the operation ; or to use water from an 
accumulator (Art. 152) in the early part of the operation, 
finishing with water from an intensifier (Art. 152). 

In an early form of COtton baling machine, instead of using 
one large ram, three were used. At first only the centre press 
was connected with the pumps, and of course it rose quickly; 
meanwhile the other two were filling with water from a tank, 
but the pressure of the water in them was insignificant. As 
the operation proceeded one of the side presses was discon- 
nected from the tank, and was fed from the pump. The 
operation proceeded more slowly now, as the pump had to 
supply two presses instead of one; but the possible total 
force was doubled. Towards the end of the operation the 
third press was disconnected from the tank, and was connected 
with the pump; the operation proceeded more slowly still, 
although the pump might be working at much the same speed 
as in the beginning. But the possible total force was just 
tliree times what it would have been with only one ram. 

In a later form there are twelve pump plungers attached 
directly to the cross-heads of the steam-engines. At the 
beginning all twelve are working, as the pressure is small. 
Then as the pressure gets greater, one set of four pumps is 
detached, so that they do not pump water into the press, but 
merely pump water back to the cistern from which they 
draw it. Thus eight pumps are now pumping, forcing into the 
press less water than the twelve did before ; but as they have 
the whole steam piston force acting on them, they are able to 
force the water in against a very much greater ram pressure. 
Later on in the operation four more pumps cease to act. 

A further improvement is to be noticed. The head or 
platten is made with two long columns attached to its under 
side, hanging down. When the first operation is finished, the 
bottoms of these columns are just above the base, and may be 
locked firmly, so that the head cannot fall back again. Now 
the finishing stroke is made; two 19-inch rams are pressed 
downwards on the upper end of the bale with a much greater 
pressure than it was possible to apply with the bottom 11 -inch 
ram. 

Other forms of cotton-press are used. Sometimes fewer 
pumps are used, and two bottom presses and rams, instead of 
one with larger rams, for pressing downwards from the top of 



APPLIED MECHANICS. 187 

the press to finish the operation. In some cases the finishing 
pressure operation may be performed when the bottom ram is 
being withdrawn, so that one bale is being finished in the 
upper part of the press when the box is being filled with cotton 
for a new bale. Diagrams have been obtained giving the 
nature of the pressures to which a cotton bale is subjected. 
In one which is before me, when the lower ram has risen 11 
feet, the force is onlj 8 tons; but in another foot the 
force increases to 16 tons; in another foot, to 33 tons; and 
in another to 59 tons. The upper rams begin to operate when 
the total force is about 160 tons; but on moving through 3 
inches, they have to exert 200 tons ; three inches further, 300 
tons ; three inches further, 500 tons ; and three inches further 
— that is, when the bale was finished — they were exerting a 
force of 900 tons. 

We see, then, that capability of working very rapidly when 
the forces are small, and working very slowly when the 
forces are great, on the supposition that the steam-engine is 
always working at the same speed — these are the important 
things to be looked for in hydraulic presses. 

152. There are hydraulic power companies now in many 
towns, whose 4- and 6-inch, and other sizes of pipes are laid along 
the streets like gas-pipes, so that any customer may be supplied 
with pressure water to work pressing, lifting, and other 
machinery, paying merely for the amount of water used. The 
force-pumps are usually worked directly from the cross-heads of 
steam-engines ; they must be capable of delivering the maximum 
supply required by customers. If the area of the pump plunger 
is a square inches, the pressure p lb. per square inch, the force 
on the plunger must he ap lbs. As there ought always to be 
a fly-wheel on the engine, it is the average efiective pressure of 
the steam, minus f rictional forces, which is determined by this. 
If the length of a stroke is I feet, and there are n efiective 
strokes per minute, the horse-power expended in the pump is 
jylan -f- 33,000. Usually there are three pump plungers working, 
so that the flow of water shall be more nearly uniform. There 
is also always an accumulator, or heavy weight w, of perhaps 
112 tons, resting on the top of a ram, of 20 inches diameter, 
its press communicating freely with the hydraulic mains. In 
such a case, the pressure in the accumulator would be 800 lbs. 
per square inch. When there is a small demand, w will be 
seen to rise, its press taking the surplus supply. When the 



/ 



188 



APPLIED MECHANICS. 




demand is great, w will be seen to fall, its press providing for 
the extra demand. Very often the engines will be seen work- 
ing fast ; w will be seen rising." When w reaches a certain 
height it acts on a lever which shuts off 
steam from the engines ; these work less 
rapidly, and w will be seen to fall. When 
w descends far enough it moves a lever, and 
admits more steam. The total store of energy 
in w lifted is often not more than half a 
minute maximum supply, yet it is sufficient 
for regulation. In the above case the maxi- 
mum store, if the maximum lift is 12 feet, 
is 112 X 2,240 X 12 foot-pounds. The same 
answer will be obtained if we remember 
that every pound of water means a store of 
2-3 X 800 foot-pounds. Or, again, every 
cubic foot means a store which is numeri- 
cally equal to the pressure in pounds per 
square foot. When, as on board ship and 
elsewhere, a great weight would be incon- 
venient, an intensifier (Fig. 126) is used. If 
water fi'om a tank at the pressure ]? communi- 
cates with one side of a piston a of area a, it 
acts just like a weight of the amount 2^ a. c 
is the accumulator, whose ram a is so much smaller than the 
piston that the pressure P in the accumulator is that necessary 
in the hydraulic mains D E. When a tank of water is not 
available for A, the pressure of steam is employed. 

Exainj)le. — If we wish to get 700 lbs. pressure from a ram 
5 inches diameter, and we have a supply from a tank 88 feet 
high, what size of piston is needed in the intensifier? Here 
p is 88 -^ 2-3, or 38-3 lbs. per square inch. (See Art. 410.) 
The total force needed is 700 x 5^ x -7854, or 13,744 lbs.; 
.so that the area of the piston, or large ram, is 13,744 -h 38*3, 
or 359-13 square inches ; or it is 21*4 inches in diameter. 

Exercise. — A press has two stuffing boxes in line, and a 
vertical rod protrudes through both, the lower end being the 
smaller. The diameters being 8 inches and 5 inches, and the 
water pressure 1,400 lbs. per square inch, what is the load on 
this differential accumulator *? If it may rise 10 feet, what is its 
store of energy, and what is its store of water in cubic feet, and 
in gallons] ^ns., 42,882 lbs.; 428,820 foot-lbs.; 2-127; 13-25. 



Fig. 126. 



APPLIED MECHANICS. 



189 



153. Pressure water, as sold by the hydraulic companies, 
would be cheaper if a better load factor (average output ^ 
maximum output from the central station) existed, as the 
engines could work continuously. They charge about 2d. per 
horse-power per hour, and the cost of production is about 0-6 
pence. The cost of the actual power given out by the motors 
used by customers varies greatly. 

Exercise. — At a pressure of 800 lbs. per square inch, 




what is the charge for 1,000 gallons of water at 2d. per 
horse-power per hour? Ans.^ 18-67 pence. 

154. A water-pressure engine may be looked upon as the in- 
verse of a reciprocating pump. If we neglect the shocks, which 
are always due to imperfect construction, when a water-presisure 
engine or pump works at a certain speed, the loss of energy by 
friction in the engine seems to be the same at high and at low 
pressure, and hence there is greater efficiency at high pressure. 
In all cases, wherever kinetic energy is produced in water, it 
is almost altogether wasted except in turbines, and to some 
extent in certain punching-machines. 

Fig. 127 shows one of the simplest forms of engine. Water 
enters the arrangement by the pipe a when the cock b is 



190 APPLIED MECHANICS. 

opened by means of the handle x. There are three rams here, 
in oscillating presses, gearing on the same crank pin ; but I 
mean to confine my attention to one of them. The supply of 
water enters at a, and finds its way to the space F, by means 
of a passage in the framework of the machine. ■ There is 
another passage leading to the exhaust spaces G, and allowing 
water to flow from these spaces through the discharge pipe Q. 
By reversing the handle, f may be made the exhaust space, 
and G the supply space, and when the handle is in the middle 
position it acts as a brake, so that by means of this handle we 
can make the engine work in opposite directions, or stop it 
altogether. 

Water enters at l, and fills the space J, and it presses on 
the ram c, causing it to leave more empty space behind it. 
You know now how to calculate the force with which the 
plimger is being pressed. The plunger cannot go out without 
turning the crank, h k. Observe, too, that as there is no 
connecting rod, the cylinder J turns just like the oscillating 
cylinder of certain steam-engines. When the crank reaches its 
dead point, the plunger can go out no further ; but when this 
happens the orifice L is just ceasing to let water enter from f, 
and is beginning to let the water escape into g. As three 
plungers act on the same crank pin, the crank does not stop 
anywhere, and as it moves on, the plunger c comes back again, 
driving the water from j through l into G and away by Q. 

This is all exceedingly simple. The quantity of water used 
in one stroke of c is simply the volume of c which leaves J in 
one stroke, assuming that the water can come in quite freely. 
Each pound of this water has a certain amount of pressure 
energy, which it gives up to the plunger, and which you may 
take to be the pressure energy in f minus the pressure energy 
in G. This is only true if we assume that the water has no 
kinetic energy where it is in contact with the plunger ; but as 
the plunger is itself moving, we know that we are here making 
a small error. This is the same as saying that the pressure on 
the plunger is less than the pressure in the mains. We are 
also assuming no loss by friction at the passage, L, which is 
again an error. 

Xeglecting friction, if p is the total pressure difi'erence 
between the supply and exhaust, and a is the area of cross 
section of the plunger, then |? a is the force upon it in pounds. 
If I is twice the lensTth of the crank in feet, and there are n 



APPLIED MECHANICS. 



191 



revolutions per minute, each plunger gives out plan -^ 33,000 
horse-power, and the three plungers give out three times this. 
Notice that this engine is almost too simple; there is con- 
siderable wire-drawing, and therefore loss of pressure due to 
friction at the valves. 

155. Fig. 128 is a section of another three-throw water- 




Fig. 128. 



pressure engine. The pistons in the three cylinders, ABC, are 
trunked and connected to one crank-pin, D, o being the centre of 
the crank-shaft. The method of packing is clearly shown. The 
water presses only on one side of each piston, and as between 
the rods and crank-pin only pushing forces act, the bearing is 
only on a small portion of the circumference of the pin. It 
can easily be imagined how the pressure water is admitted to 
each cylinder at the end of the stroke of its piston, and how it 
is allowed to escape at the other end. The engine will start 



192 APPLIED MECHANICS. 

from any position, and the flow of water to it is fairly regular, 
because three cylinders are used. 

156. It is evident to anyone who has studied air- and 
steam-engines that water-pressure engines may be of a very 
great number of shapes. 

In most engines of this kind the work to be done per stroke 
may be very different at different times, and yet the pressure 
water used — that is, the energy — is always the same, and so there 
is considerable loss. One method for remedying this evil is to 
shorten the crank as the work being done is less. 

Another method which has been suggested is that of 
admitting pressure water for less than the whole stroke, siniply 
taking water from the discharge-pipe for the remainder. When 
engines have a fixed sort of duty, there is no need for any 
adjustment. 

The common construction of water-pressure engines will be 
readily understood if you understand the construction of the 
steam-engine. Remember, however, that the velocity of water 
ought never to be great in the engine or pipes. Wire-drawing 
leads to serious loss by friction in the steam-engine ; it is far 
more serious in water-pressure engines. In these the valves 
ought to be quite open, giving a very large passage for water 
to flow through almost immediately. Hence, although the slide 
arrangements of Fig. 127 are allowable in small engines, they 
cannot be used in large economical engines working constantly. 
Remember, too, that all frictional losses are made much greater 
by quick motion, and by reversals of motion, and hence that it 
is very important to have a long stroke of piston or plunger. 

Lastly, remember that, although there ought to be no waste 
space between steam-piston and cylinder at the end of the 
stroke (very little clearance), this is of almost no importance 
in water-pressure engines, because of the incompressibility of 
water. 

157. We have described (Art. 136) the force-pump used by 
hydraulic engineers. All force-pumps are much the same in 
principle, but in ordinary force-pumps for water (see Art. 406), 
and for use in feeding boilers there is usually less trouble with 
the valves and packing than there is in hydraulic pumps. 
Students will pay attention in Chap, XXIII. to the blow 
caused by sudden stoppage of the flow of water. 

Direct-acting steam-pumps are much used in draining mines. 
Sometimes water-pressure engines have been used instead of 



APfLtED MECHANICS. l93 

steam — that is, water from an accumulator on the surface goes 
down the mme to a water-pressure engine which drives a pump; 
the mine water and the exhaust water of the engine are sent 
to the surface. In such a case, pressures are very great, and 
the effects due to sudden stoppage of flow in pipes may pi-oduco 
damage unless the valves are so arranged that when they close 
or open the general circulation is not much affected. 

The common lift-pump is described in Art. 406. Its 
form as an air-pump is usually described in books on 
the steam-engine, and, indeed, it is in such books that 
pumping machinery more naturally finds a place than in 
books on applied mechanics. In Art; 433 we find that in 
a stream line the total energy of one pound of fluid remains 
constant if there is no friction ; a pump adds to this store ; a 
turbine or water-wheel takes away from this store ; friction 
also diminishes the store, convei'ting mechanical energy into 
heat. 

158. What now are the conditions under Avhich trans- 
mission of power by hydraulic action is most suitable ? 

1st. Intermittent action, because the accumulator is so 
nearly perfect, giving out energy simply in proportion to the 
quantity of water used, and yet allowing an engine of small 
power to be storing continually. 

2nd. Action requiring not a very great quantity of power. 

3rd. Action of a comparatively slow kind, the water never 
being allowed to fi.ow so fast that its store. of kinetic energy is 
great, since the kinetic energy is nearly all wasted ; slow action, 
with considerable force. 

4th. Action which is greatly continuous in one direction, 
not requiring much stoppage or reversal of the water motion. 

You will see from this that the conditions required in 
pressing machinery, cranes, hoists, and lifts are better satisfied 
by hydraulic transmission of power than they can be by any 
other method of power transmission which is known to us. 

159. In Fig. 129 we have one specimen of a hydraulic crane, 
whose action it is very easy to understand. Suppose water at 
700 lbs. pressure per square inch admitted to the cylindric 
space A, and that the space B, on the other side of the piston, 
although filled with w^ater, has only a comparatively small 
pressure, and communicates with a low-lying tank ; neglecting 
the small pressure in b, we see that the piston is pushed for- 
ward with a force of 700 x A lbs. if A is the area of the piston 

H 



194 



APPLIED MECHANICS. 



in square inches. Now the motion of the piston is multiplied 
eight times by the chain, which passes over blocks, each contain- 
ing four sheaves, attached at M and N. The block at x gets 
the motion of the piston, and the chain at p must be drawn in 
eight times more quickly than this. You know that the pull 
in the chain may be one-eighth as much as the total force on 
the piston, and it can therefore lift through eight times the 
distance a weight of one-eighth the amount, or 700 a -^ 8 lbs. 




Fig. 



This is the original form of Lord Armstrong. Whatever 
defect there is lies in the use of chains passing over numerous 
sheaves giving rise to a great amount of friction. Cranes require 
so little horse-power to work them, however, that mere economy 
of coal is barely worth considering, and the risk of accident, which 
might be done away with very greatly by direct hydraulic action, 
is not important either. You see that if a and b both receive 
pressure water there is less water used than before; for although 
as much comes into A as before, b is sending water back to 
the pump or into a. In fact, the total pressure on the piston 
is 700 lbs. multiplied by the difference between the areas of the 
two sides of the piston exposed to pressure — that is, the mere 
area of cross-section of the thick piston-rod. Hence, we can work 



APPLIED MECHANICS. 



195 



this crane so that it lifts heavy loads or light loads — that is, it is 
double-powered. Unfortunately, however, when working any 




Fig. 130 



heavy load, it is consuming as much energy as if it were lifting 
the heaviest load it is capable of lifting. When lifting on its 
second power, and lifting a light load, it is using as mucii energy 



196 APPLIED MECHANICS. 

as if it were lifting the heaviest load this second power is 
capable of lifting. 

Thus, let us suppose it lifting eight tons to a height of 20 ft., 
and that one cubic foot of water is used in the operation. Then, 
on the same power, suppose it to be lifting four tons to the same 
height, it still uses one cubic foot of water, and in both cases 
there is the same energy used for the pumps. Of course, by* 
a combination of cylinders, it would be possible to vary the 
work expended as the load varied, but the expedient is not 
of a very practical character. 

Such a crane as this has usually another cylinder and chain 
for turning it round. In many cranes the action is more direct ; 
a two-sheaved pulley-block, or one with a single sheave, or 
even no pulley-block at all, being used. (See Fig. 130.) In 
many cases loads are pushed up on the top of a ram, in 
others they are pulled up by a piston rod directly ; in fact, 
there is an endless variety in the arrangements, but the 
principles of the hydraulic press parts are the same in all. 

160, Now, supposing it is easy to get a supply of pressure- 
water, what, besides cranes, hoists, and pressure-engines, can 
be worked by means of it ? All forging and welding machines, 
which with moulds and dies properly shaped and pressed 
together with enormous steady force, seem, for objects of 
settled shapes, to be a very great improvement on any method 
of hammering; stamping machines, for all sorts of purposes; 
and bending machines, for joggling and bending angle-irons, 
rails, and beams. 

Students must examine such bending-machines for them- 
selves, and notice how the travel of the press-block is deter- 
mined by tappets, which open and close the valve for water 
supply at any point in the stroke we please. By means of such 
a bending-machine any number of curves may be made identi- 
cally the same. In many punching, forging, stamping, and 
shearing machines we also have the inlet and exhaust valve 
worked by hand or foot levers, the stroke being regulated 
with great nicety by tappet motions. Thus, in a riveter, a short 
ram of eight inches in diameter may be used. Used with an 
accumulator and pumps of its own, the pressure is usually 
1,400 lbs. per square inch ; so that the total force available is 
31 tons. 

Next we have tools which are easily portable. In Fig. 131 
the ram gets a motion about its centre, its axis and line of actior 



APPLIED MECHANICS. 



197 



being an arc of a circle. By this compact arrangement we do 
away with connecting-rods and many other complications, and 
we get a great increase in stiffness. Instead, then, of bringing 
work weighing many tons to a machine, we bring a little 
machine, weighing 5 cwt,, to the work ; we can punch holes 
and finish the riveting ; that is, we can finish most of it — for 
tliere must always remain rivets in difficult positions which 




Fig. 131. 



require to be done by hand — with no other intermediate 
gearing between this little riveter and the steam-engine than a 
small jointed pipe, the two forms of the universal joint which 
are used being made water-tight by leather collars. Instead of 
a jointed pipe, a pipe formed into a spiral forms a good yielding 
connection. It is, perhaps, in looking at these little riveters, 
rather than in any other examples, that you will be struck by 
tiie simplicity of hydraulic working. The flexible pipe trans- 
mits power more faithfully than huge beams and cog-wheels 
would do. 



198 



APPLIED MECHAXICS. 



161. The machines which I have described are well suited 
to hydraulic work. A punching, or shearing, or stamping, or 
riveting machine, driven by shafting, repeats its stroke at regular 
intervals, and the workman cannot arrest a stroke. He wastes 
time in waiting for a stroke, but when a stroke is being made, 
and he sees that his plate has been wrongly placed, or has shifted 
its position, he must just as patiently watch the inevitable 



Irorv Bar 
0'78 in/, thicky 

_ fOCO' 





tsoo- 


r22-urh.hol^. \ 
yiSuvplcde / 
Iron.' / 


V 1COO - 




\ 600- 
9— 


f 


r 



-C 



Fig. 132. 



Fi-. 133. 



^nWOXJbs persq iriz-J^y 
cnv accanrax 
lbs 



sq ijn/-K 




Fig. 134. 

completion of the stroke. With the hydraulic punching- 
machine, on the contrary, the workman can stop the motion 
at any instant, even if the punch has made its mark on the 
surface of the plate. He can start instantly from a condition 
of rest ; he has not to wait till he pulls the belt on to the fast 
pulley, or starts the donkey-engine, and he knows that when 
there is no stroke being made there is no power being wasted. 
Then, again, all the shafting and other machinery of a large 
shop have not to be set in motion, or varied in their motion, 
to punch a five-eighth inch hole. A man wastes just the same 
energy in punching this one hole as if it were one of a hundred 



APPLIED MECHANICS. 199 

he was punching. Suppose, again, that a man carelessly puts 
a 1 J-inch plate under a punch arranged for a five-eighth inch 
plate. The sudden blow of an ordinary punching-raachine, 
with its fly-wheel, would produce a fracture somewhere. In a 
hydraulic machine there is a simple stoppage. Think, too, of 
the strength of roof and columns needed to carry shafting ; of 
the trouble in the use of overhead cranes when there are 
many shafts and belts ; and, above all, think of the noise, in 
comparison with the invisibility of pressure mains, and the 
dead silence of hydraulic tools. Observe, too, that these 
hydraulic machines need but little foundation. 

The diagrams of Figs. 132, 133, and 134 (p. 198), showing the 
water-pressure in the cylinders of various tools at Toulon during 
their stroke, are exceedingly interesting. Fig. 132 shows a 
curve from a punching-machine. The pressure is not equal to 
that in the accumulator, unless the tool has so much resistance 
to overcome that it is moving very slowly indeed. The sudden 
rise shows what occurs when the punch is just beginning to act 
on the plate. In practice you must understand that this early 
part of the diagram has no existence ; Mr. Tweddell's tappet 
motion prevents such waste. The area of each diagram is 
roughly the amount of energy utilised. The area of the rect- 
angle in Fig. 134 shows the energy taken from the accumu- 
lator. In all these diagrams, then, it must appear to you that 
there is a large amount of waste. This is greatly reduced by 
the tappet motion, and in any case it is very much greater in 
such tools when driven by shafting. 

Fig. 133 shows a diagram from a plate-shearing machine. 
The uniformity of the pressure is due to the angle which the 
edges of the shears make with one another. 

The riveting-machine diagrams are most interesting. 
Neglecting the useless part of the stroke in Fig. 134, we see 
the rise of pressure, e f, due to the setting up of the rivet, f h, 
the clinching of the rivet and closing of the plates. The sudden 
stoppage of motion of the water gives a blow, shown by the 
pressure becoming half as much again as the accumulator 
pressure, and we rely on this blow as perfecting the filling of 
all cavities. 

162. In Fig. 135 two rams, A and <x, are equal, one carrying 
a gun which is to be quickly lifted and lowered, the other a a 
counterweight. As it is not convenient to vary the counter- 
weight, and as there is really balance only in one position (for 



200 



APPLIED MECHANICS. 



na 



p 



either A or a getting lower causes the pressure acting on it 
upwards to get greater), a plan taken is this :— a and a have 

two presses communicating 
by means of a small pipe; 
and E, which is greater than 
R, is partly supported on a 
small extra ram of area a' in 
a press of its own, which may 
either be made to communi- 
cate with the other two 
l^resses, or with a neighbour- 
ing tank where the pressure 
is small. In the one case, 
E rests on what is practic- 
ally a ram of area, a + a', 
Fig. 135. and it is lifted ; in the other 

case on a ram of area a, and 
it falls. The supply of pressure water to a' is effected by 
uneans of a pump and accumulator. When piessures are 
already very great, change of pressure due to rise and fall of 
a ram is not very important, but in long rams it is often 
important. 

163. In most of the machines which we have described, 
although M^ater usually changes its level during the action, 
this change of level has been so small as to be negligible. 
But in nearly all lifting operations we have to consider the 
work done in the lifting of water as the ram rises. 

We all know the conditions required in an ordinary hotel or 
chambers hoist; those conditions are absolutely the same for 
warehouse hoists, because a hoist which carries goods occasion- 
ally carries men in charge of these goods. Long ago, I had 
some designing and carrying out of mill hoists, in which the 
cage was lifted by a rope passing over an elevated pulley, 
driven from the main shafting of the mill, and stop})ed at any 
point of ascent or descent by automatic disengaging apparatus 
which also braked the pulley. The cage was balanced by 
counter- weights, as a window is balanced. Our greatest trouble 
was in the arrangement of safety apparatus, which would stop 
the cage in falling should the rope break. ISTow, it is well 
knowm that such safety apparatus can never be thoroughly 
depended upon, however ingenious its design may be, because 
the ordinary working of the hoist does not keep the safety 



APPLIED MECHANICS. 



201 



apparatus iii action ; immunity from accidents causes it to be 
neglected, and when an accident does happen it will not work. 

There is nothing so safe as a hoist whose rapid motion is 
resisted by a considerable amount of friction. But, unfor- 
tunately, if the friction is that of solids on one another, there 
is as much frictional resistance to the ordinary working of 
the hoist as there is when an accident occurs, and hence 
assurance of safety by friction means tremendous loss of 
power at all times. 

Now, you remember that the frictional resistance of water 
was of quite a different kind. There is almost no resistance 
to the flow of water, if the flow is slow. There is only a 
moderate loss of power in the ordinary use of a hydraulic hoist; 
but the motion cannot become too rapid for safety, for the 
frictional resistance is exceedingly great at high speeds. 

Although, therefore, serious accidents cannot bb wholly 
prevented — water may leak away by valves and leave the cage 
wholly unsupported, for example — yet there is more safety 
possible with hydraulic than with other 
hoists. 

164. In a great many hydraulic 
hoists the action is precisely the same as 
in Armstrong's cranes. Fig. 136 shows 
such a construction, used by Armstrong 
himself. A is a pressure cylinder, with 
its ram carrying at B the movable 
block with sheaves, which pull the 
chain or wire rope, m, k. There is a loss 
of effect, due to the altering weight of 
the chain, as the cage rises or falls. 
This difficulty may be got rid of by 
letting the ram move vertically, when 
the altering weight of the ram itself 
may be made to balance the alter- 
ing weight of the chain. All such 
hoists as this can be readily balanced, 
so that, the dead weights may balance 
at all points in the ascent and de- 
scent. They are, however, subject 
to the risks inseparable from the use 

of chains or ropes, and must be regarded as unsatisfactory 
for this reason. That the lifting of every load means the 




(c9 



£ 



^^ 




Fig. 136. 



202 APPLIED MECHANICS. 

expenditure of the same amount of energy is not a considera 
tion of any importance in these hotel hoists. Of course, 
there is a slightly greater speed when the load is small, as 
the water pressure is capable of lifting the heaviest probable 
loads ; but you know enough already about water friction to 
see that the increase of speed must be insignificant. This con- 
dition is the same for all hydraulic hoists hitherto constiucted. 
Very often we use a direct acting hoist. Here the ram moves, 
pushing the cage up directly. When the pressure of water is 
very considerable, say 200 lbs. per square inch, and the lift is 
not too high, this form of hoist is good; for although rather 
wasteful of power^ it is exceedingly simple. The press is 
sunk so far beneath the basement that there is room for 
the whole length of the ram when the cage is in its lowest 
position. 

165. It is necessary now to consider the diminution of 
lifting force as a ram rises, and we return to Fig. 135. 

Let the total load upon the ram of area a, including its 
own weight, be called e, and let the total load on the ram of 
area a, including its own weight, be called e, and for easier 
calculation assume the ends of the rams to be flat and horizontal. 
When a is just about to descend, let the end of a be A feet 
above a. When e falls one foot we found how high r must rise 
if we could only neglect the weights of the water. But we shall 
now consider the weight of the water, and take the areas a 
and A to be in square feet, and the forces R and e to be in 
pounds. 

When a falls a very short distance, x feet, further displacing 
ax cubic feet of water, or 62 "3 ax lbs. of water, this water is 

lifted through h feet. Now A will rise cc x — feet, and the 

A 

work done by e in falling being E x, is equal to the work 
R X a? - done in lifting E, and also to 62*3 a x Ii, the work 

A 

done in lifting the water. Writing this down algebraically, 
we find — 

or the pressure at the end of a, Avhich is e -f- a, is greater than 
that at A, which is u -^ A, by the amount 62-3 h lbs. per square 
foot. We find this to be an increase of 2,116 lbs. per square foot, 



APPLIED MECHANICS. 



203 






1 



or 14 '7 lbs. per square inch, or one atmosphere as it is called, 
for every 34 feet of difference of level. 

When a ram rises 34 feet, if the 
supply pressure to the press remains the 
same, the lifting force on the ram 
diminislies 14-7 lbs. per square inch^ 
and for a 68 feet rise the diminution 
of pressure would be twice as much. 
These changes of pressure are not very 
important when we are dealing with 
pressures in the press of 200 or 300 lbs. 
per square inch, but they may be very 
important at lower press pressures, 
and in some cases we make the supply 
pressure increase as the ram rises. 
Notice that you may look upon the phe- 
nomenon in two seemingly different 
ways. You may either say to yourself, 
" As a stone is lighter when sur- 
rounded by water, so this ram is lighter 
when it is at the bottom, for more of it 
is surrounded by water in the press " ; 
or you may put it in this form, "The 
pressure on the bottom of the ram must 
just balance the weight of ram, cage, 
etc., but as the bottom of the ram rises, 
this means that we ought to have a con- 
stant pressure at the bottom of the ram 
wherever it may be, and consequently 
a gradually increasing pressure in the 
cylinder everywhere as the ram rises." 
Now, I do not care which of these two 
views you take, but you must not mix 
them, and say that "not only does the 
ram get heavier, but it needs a greater 
pressure at its lower end as its lower 
end rises." I prefer always to say, 
"The ram appears to get heavier just 
in proportion to the amount it has 
been raised, and this must be balanced 
by increasing the pressure of the 
supply water." Fig^g^^ 



204 



APPLIED MECHANICS. 





Now, remember tbat our supply water in the direct- 
actiug hoist is at a constant pressure, 
and you ^s'ill see that it is quite 
impossible, with such a simple 
an-angement, to have perfect uniform- 
ity of action, although it is approxi- 
mated to more and more nearly as the 
pressure is greater. In this kind of 
hoist it is usual to let the water escape 
from the cylinder to a discharge cistern 
considerably abo^"e the cylinder, so that 
in its descent the ram and cage may 
not fall too rapidly. Here, again, we 
have the same want of uniformity of 
action, since the apparent weight of 
the ram gets less as it falls. 

166. The usual practice has been to 
nearly balance the dead-weight of ram 
and cage by a weight, as in Fig. 137 
(p. 203), so as not to require too high 
a Kft in the discharge-pipe, and to so 
arrange that the varying weight of chain 
shall just balance the apparent change of 
weight of the ram. It is evident that 
if the ram rises one foot, the counter- 
weight, increases by the weight of two 
feet of chain ; hence, the weight of two 
feet of the chain ought to be equal to 
that of water occupying the volume of 
one foot length of the ram. I7nfortn- 
nately, these chains and counterweights 
destroy the simplicity and absolute 
safety of the hydraulic hoist. If the 
ram were to break near its upper end, 
the cage would be drawn violently 
upward by the chain. The upper part 
of the ram is in tension, and the lower 
part in compression. 

It is obAT.ous, then, that there must, 
for a complete and perfect hydraulic 
lift, be such a regulation of the pressure 
of the water as it enters the cylinder 



Fig. 138 



APPLIED MECHANICS. 205 

of a hoist that the only force to be overcome shall be the 
variable weight placed in the cage, whether that of passengers 
or goods, together with the necessary friction. The hydraulic 
balance hoist satisfies this condition. It can be worked with 
either high or low pressure water ; the ram is always in com- 
pression, supporting the load, and no part of the machinery is 
above the cage, and there is no part of the machinery likely to 
break in such a way as to cause an accident. 

This hydraulic balance lift is shown in Fig. 138 (p. 
204). The hydraulic cylinder, ram, and cage are as usually 
made, except that the ram is somewhat smaller in diameter. 
Its size is determined by the strength required to carry the 
load, and not by the working pressure of water available. The 
lift cylinder is in hydraulic connection with a second and 
shorter cylinder, e, below which there is a cylinder, c, of larger 
dimensions. There is a piston in each, connected by the rod. 
The capacity of the annular space e below the upper piston 
is equal to the displacement of the lift ram. The annular area 
of the lower piston is sufficient, when subject to the working 
pressure of the supply water, to overcome friction and lift the 
net load ; and the full area of the upper piston is sufficient, 
when subjected to the same pressure, to balance within a small 
amount the unalterable weight of the ram and cage. 

Assuming the cage at the bottom of its stroke, the valve 
is opened by a man in the cage pulling on a rope, by a system 
of levers, and pressure water is admitted. The pressures on 
the two pistons cause them to descend, forcing water from the 
annular space to the hoist cylinder. The hoist ram ascends, 
and in doing so gets heavier, but the pistons are descending, 
and the total pressure on them is getting greater just in the same 
proportion. When the ram reaches the top of its stroke the 
valve is closed and the lift stops. Now open the exhaust valve, 
which lets the water pass away from c — only from c, re- 
member — and the weight of the ram and cage presses the 
water from the lift press into E, causing the pistons to rise. 

To make good any possible leakage, provision is made for 
admitting pressure water under r, and so raising it, the lift 
ram being at the bottom of its stroke, that water will flow into 
the space. 

167. We see that the hydraulic hoist has — (1) The great 
element of safety from the absence of possible breakage of chains 
or ropes. But in some forms it is not without a danger of its 



206 APPLIED MECHANICS. 

own — namely, the danger that when the cage is remaining 
caught in a fixed elevated position for a time, the cylinder 
may be emptying of water through a leakage of the valves. 
(2) That the expenditure of energy depends very little 
upon the dead load. But there is still the drawback that 
every load, however small, requires the same expenditure of 
energy as the greatest load which the hoist can lift. This 
drawback is common to all hydraulic hoists such as I have 
been describing. 

168. You can now have no difficulty in understanding the 
construction of all warehouse and hotel lifts. The hydraulic 
principles involved in all lifts are the same, but with larger 
weights to be raised there are peculiarities of construction 
which ought to be studied from actual drawings of such lifts, 
as, for example, the pair of lifts at the Seacombe Pier, on the 
Mersey, to take carts and waggons from the floating landing- 
stage to the high level. Tiie height of lift is 32 feet, and the 
net load 20 tons. Here we have simply direct acting rams and 
presses sunk in the river-bed, the presses being surrounded by 
cast-iron protecting cylinders. In this case there is no attempt 
to balance the increasing weight — that is, the displacement of 
the ram as it rises. Each cage, or platform, is supported on 
one ram, and the designer has to take care that the weight of 
any waggon shall be so carried on the top of the ram that no 
part of the structure is unduly strained. The waggon rests on 
a platform, which can slide on the bottom of the cage, so that 
although the cage rises vertically, the platform may everywhere 
be in the position in which it would be on the landing-stage, if 
the landing-stage rose and fell. The landing-stage is attached, 
as we know, by girders, which do not alter in length, so that 
it does not rise and fall vertically, but really in arcs of circles. 
In this case, the lift is of a variable amount, depending on the 
height of the tide. There is a connecting valve between the 
two presses, so that a descending load in one lift may raise a 
less load in the other when necessary. 

169. This idea of having two lifts side by side, so that the 
lowering of one may cause the rise of the ether, you will under- 
stand better in the original example of its use, the canal lift 
of Messrs. Clark and Standiield, on the River Weaver, in 
Cheshire. Figs. 139 and 140 show the canal and the river, 
one 50 feet above the other. We want to raise or lower canal 
boats from the one to the other, and we want to avoid the 



APPLIED MECHANICS. 



207 



expenditure of water, and the delay which occurs when there 
is a chain of locks. 

There are two great wrought-iron troughs, each 75 feet long 




Fig. 139. 



and 15 J feet wide, each being carried on the top of a ram 3 feet 
in diameter. Now, when I tell you this, you will at once see 
how incomplete my description is. Each tremendous trough 



208 



APPLIED MECHAXICS. 



is carried easily by one ram acting at its centre. You can, in 
your imagination, go into all the details of girder- work necessary 
for the safe carrying of such a load in such a manner. The 
weight of each trough, with the water and barges in it, is 240 
tons, and this gives a pressure of about one quarter of a ton 
per square inch in the press. At each end of each trough there 
is a gate. 

When the trough is up, and gate A is lifted, the trough 
forms part of the aqueduct. A barge floats into it from the 
canal, and the gate is closed again ; also 
the aqueduct end is itself closed with a 
gate. Now the trough is lowered con- 
taining the barg'e, and when it reaches 
the lower level, gate b is lifted, when, 
of course, the trough really forms part 
of the river. You must describe to your- 
selves how this press is sunk — how we 
have a tunnel which enables us to 
examine the packing of the cylinders, 
how these great columns are firmly sup- 
ported, so that we may have guides to 
prevent the tilting of the troughs. All 
this is constructive detail which you can 
read about in the Proceedings of the 
Institution of Civil Engineers. You 
might fear also that the joints of the 
gates at the end of the trough might be 
leaky, and, above all, that the joint of the trough with the 
aqueduct end might be leaky ; but even from the figure you 
can see how perfectly these difficulties can be got over. 

Suppose I have a boat in any vessel of water, and 1 know 
that the water is at a certain level, and I take out this boat 
and put in another boat, and add water or take it away until 
the level is just what it was before. You know that the weight 
of boat and water is always the same. The total weight simply 
depends on the level of the water. The weight of the boat 
alone is always equal to that of the water it displaces, and, 
therefore, the trough filled to a certain height with water, if 
there is no boat, will just weigh the same as if there is a boat 
in it and the water is at the same level as before. 

Now, suppose that the trough m is part of the canal, and 
there is or is not a barge in it, and suppose that N is down, 




Fig. 140. 



APPLIED MECHANICS. 209 

and is in communication with the river, and there is or is 
not a barge in it. Now close the gates. Suppose there 
is five feet of water in m, and that, if there is not four feet 
six inches of water in N, we let water into or out of it from 
the river till the water is at this level. Now let the valve be 
opened ; water will flow from one press to the other, for M 
is heavier than N, and m will fall, causing n to rise. Suppose 
the lifts were very high indeed, it is evident that this fall- 
ing w^ill not stop till m is w^ell below the level of N ; in 
fact, till the lightening of the ram m, as it displaces more 
water, and the increased weight of ram n as it leaves the water 
in its press — till these two added together are equal to the real 
difference in the loads of m and n. In the present case, the lift 
is not great enough for such an effect to occur. But as soon as 
M enters the river it gets very much lighter in weight. Indeed, 
as soon as it sinks six inches into the river, the weights of m 
and N are equal, so that leaving out of account the displace- 
ments of the rams, N cannot be raised any further by the falling 
of M. There is exact balance, and no further motion, then, 
when N and m have still more than four feet six inches further 
to travel. The valve is now closed ; there is no further com- 
munication between the presses. The press of n is put in 
communication with an accumulator, which lifts n, pressing it 
home at its water-tight joint with the aqueduct. The water in 
N is six inches lower in level than it ought to be, however. 
Water is allowed to pass from the aqueduct through a valve 
into the space between the two gates of the trough and of the 
aqueduct, when the latter are easily lifted ; the former are also 
easily lifted now, and water passes freely into n, as it is now 
part of the aqueduct, and the barges can be taken off into the 
canal. Meanwhile we have left m still resting in great part 
on its ram. The water from its press is now allowed to 
escape, and M sinks in the river, its river-gate is lifted, and 
the barge is taken out. 

Twelve siphons in M were put in working order during its 
immersion, and when it is being lifted again they empty it 
down to the level of their free ends, which are adjusted to leave 
exactly four feet six inches of water in m. Thus there is an 
automatic adjustment of the water-levels in m and N, so that 
whatever be the weights of the floating barges they contain, 
the total weight is always the same for every operation. 

The waste in an operation comprises, first, six inches depth 



210 APPLIED MECHANICS. 

of water in one of the troughs. The falHng of this 50 feet is 
1,800,000 foot-pounds. Secondly, the accumuktor raises xfour 
feet six inches, and as the weight of n is about 240 tons, the 
waste here is 2,419,200 foot-pounds. The total waste is, then, 
about 4,219,200 foot-pounds. 

ISTow, suppose the same operation were performed down a 
flight of ordinary canal locks, under favourable circumstances, 
they would require that 14| feet depth of water of the area of 
one lock should fall 50 feet, or there \^^ould be an expenditure 
of 51,500,000 foot-pounds, or twelve times as much as with the 
lift. If, however, a canal has a plentiful supply of water, this 
is not of such great importance as it seems. It becomes of 
great imjDortance when, as in the present case, there is but a 
small supplyof water. The advantage of such a system as the 
present is rather, in my opinion, in the fact that the operation 
is finished in eight minutes, whereas in a similar ascent or de- 
scent, but by means of a flight of locks, at Runcorn, the opera- 
tion of letting one barge through requires one hour and a half. 

170. We are, however, more interested, just now, in the 
hydraulic question, the saving of energy, the saving of water 
from the canal, and accumulator energy, and so we may consider 
a somewhat similar canal lift which has been constructed at 
Fontinette, Belgium, to replace a flight of five locks, with a total 
fall of 43 feet. The troughs are of double the length, and are of 
greater depth than the last. Each ram is six feet six inches in 
diameter, and there is the improvement that we have what are 
called compensating reservoirs. Water flows from one of them 
to the descending trough, thus increasing the weight of it, just 
in proportion as its ram becomes immersed in its press ; and 
water flows back again from the ascending trough to the 
reservoir, just in proportion as its ram comes out of its press. 
Thus the ram displacement is balanced. But there is a further 
improvement : the descending trough does not descend into 
water, for this made it get light too soon ; it descends into a 
dry chamber, and only becomes a portion of the lower canal 
when the gates are lifted. Thus the falling of one trough 
can lift the other all the way to the upper level, and the 
accumulator is only needed to suppl}^ leakage from the presses. 
A single operation causes a loss of 20 tons of water from the 
upper level, or less than 2,000,000 foot-pounds of energy 
altogether, and yet the troughs have more than twice the 
capacity of those at Anderton. 



APPLIED MECHANICS. 211 

171. We have noticed here one of the methods adopted for 
balancing the change due to displacement of rams. But 
many other methods have been adopted to suit special cases, 
in which the falling of an accumulator ram causes the rise 
of another ram. Thus, for example, we let a ram rise vertically 
above the accumulator, passing water-tight into a tank of water. 
Its increased apparent weight (it is always covered), as the 
accumulator ram descends, compensates for the displacement 
effects of all the other rams in the arrangement. Again, we 
may have a tank of water as part of the load of the accumu- 
lator, and its water-level is kept the same as that of a 
neighbouring fixed tank, by means of a siphon, so that the 
weight increases as the accumulator ram sinks. By properly 
shaping this tank we are able to get any variation of pressure 
that is wanted during an operation, for such purposes as cotton- 
pressing, etc. It is easy to see that we have here a means of 
balancing accurately the dead weight of any platform and its 
ram, which has to be raised and lowered, at every j)oint during 
an operation, by the use of an ordinary accumulator. 

172. Another very important improvement in lifts is this. 
Suppose that a bridge has to be lifted by rams at its ends. 
Suppose the presses of these rams are connected with an ac- 
cumulator, they are not likely to rise equally fast, and the 
bridge would be tilted ; indeed, one of them may not rise at 
all, aud the other may be rising quickly. Again, suppose 
we have a separate accumulator for each separate press, so 
that there is no water communication between them, great 
care has to be taken at the valves to make the ends of 
the bridge rise equally fast. This difficulty is got over by 
having two presses and two rams on the accumulator. Now, 
let each accumulator press be connected with each bridge press. 
The frame carrying the accumulator weight is guided so that 
it cannot tilt ; each equal ram, therefore, falls the same amount 
as tlie other, and the equal bridge rams must rise one as much 
as the other. 

When a large structure has to be lifted, the lifting presses 
^ which get this synchronous action are so distributed in sup- 
porting the structure that there is no tilting of it possible. 
1 Fig. 141 shows an hydraulic grid. The strong girder, 
with its projecting ribs, rests on the ends of a number of rams, 
j whose presses are sunk in the bed of the dock or tidal river. 
The keel of a vessel is brought directly over it, and secured in 



212 



APPLIED MECHANICS. 



position by the bilge blocks, and side -shoring frames; the 
presses are worked, and the raras lift the grid and the vessel 
above the level of high water. Now a number of struts, which 
were previously held up horizontally, are liberated, and haug 
from the grid alongside the raras ; on lowering the rams, the 
lower ends of these struts fit into the tops of the presses, 




Fig. 141. 

forming a support for the grid, and the rams are withdrawn 
into their presses. 

There are only a few guide columns needed for the grid to 
slide against as it rises and falls, because the presses are 
arranged in three equal groups on the above-mentioned princi- 
ple, supplemented by an automatic safety valve, which lets the 
water escape from a press when its ram has risen more than 
the others. Hence the grid must, when rising and falling, 
remain perfectly level. 



APPLIED MECHANICS, 213 

EXEECISES. 

1. Hj-^draulic jack; velocity ratio of lever, 30; ram, 2|- inches 
diameter ; pump-plunger, f inch diameter. If it is experimentally found 
that, E being effort on handle, w the weight lifted, there is a straight-line 
law connecting e and w ; and if, when w = 1,605 Ihs., e = 10 lbs. ; when 
w = 6,805 Ihs., E =-- 50 lbs. ; show that w = 305 -f 130 e. If when 
w = 7,000 lbs. the pressure of the water is found by a pressure-gauge to 
be 1,932 lbs. per square inch, what is the loss by friction at the leather? 
Assuming the same percentage loss at the two leathers, what is the law 
sonnecting e and the force p with which the lever acts on the plunger ? 

Ans., 8-9 per cent. ; p = 40-5 -j- 17-3 e. 

2. A steel hydraulic press 13 inches internal diameter is 3 inches 
thick. What is the greatest tensile stress when there is a fluid pressure 
of 2| tons per square inch ? Ans., 7 "6 tons per square inch. 

If a steel pipe is 1 inch in diameter inside, and the greatest tensile stress 
is to be 5 tons per square inch when there is a fluid pressure of 3 tons per 
square inch, what is the thickness of the metal ? Ans., ~ inch. 

3. In an accumulator the average pressure is to be 700 lbs. per square 
inch; ram, 12 inches diameter. What is the necessary weight ? If the 
ram rises 10 feet, what energy is stored up ? Neglect change of pressm-e 
due to lifting. If the pressm-e is found to fluctuate between 720 and 
680 lbs. per square inch between slow lifting and slow falling of the 
weight, what is the force of friction ? If the pressures are settled for a 
position half-way up, what is the real fluctuation, taking friction and 
change of level into account ? 

Ans., 79,200 lbs. ; 792,000 ft.-lbs. ; 2,260 lbs. ; 2,505 lbs. 

4. In a three-cylinder single-acting pressure engine like that shown in 
Fig. 128 each piston is 4 inches diameter and 3 inch stroke ; the average 
acting pressure is 700 lbs. per square inch. What is the indicated horse- 
power at 50 revolutions per minute, assuming an average back-pressure 
due to friction, etc., of 200 lbs. per square inch ? The coil of a rope on a 
drum on the crank-shaft is 18 inches diameter ; what is the average puU 
in the rope if the brake horse-power is 0*7 of the indicated ? If the pull 
is to be only 300 lbs., what ought the stroke to be altered to ? 

Ans., 7-14; 700 lbs. 1-3 inch. 

5. The area of the piston of a hydraulic crane is 90 square inches on 
one side and 40 on the other; it pushes a three-sheave ptdley-block. If 
the water-pressure is 700 lbs. per square inch, what weights can be lifted 
— (1) when pressui'e-water is admitted on one side only, (2) when ad- 
mitted on both sides. Take the efiiciency of the hydraulic parts as 0-9, 
and of the pulley and crane parts as 0-4. If the full loads in the two 
kinds of working are being lifted, what work is done per cubic foot, per 
pound, and per gallon of water? 

Ans., 2,100 lbs. ; 3,780 lbs. ; 36,290 ft.-lbs. : 582-6 ; 5,826 ft.-lbs. 

6. A hydraulic punch has a ram 8 inches diameter ; |-inch holes are 
being punched, each requiring a force of 70,000 lbs. What is the water- 
pressure ? This water comes from a steam intensifier ; the area on which 
the steam acts is 300 square inches ; the area of the ram is 30 square 
inches. What is the pressure difference of the steamj neglecting friction ? 

Ans., 1,393 lbs. per square inch ; 139 '3 lbs. per square inch. 

7. In a pipe from which a press is supjDlied, the pressure is in one case 
400 lbs. per square inch and in another it is 100 lbs. per square inch ; the 



u 



APPLIED MECHANICS. 



end of the ram of a hoist is 70 feet below the level of the pipe, and 
gradually rises to be 10 feet above the pipe. AVhat is the change of 
pressure ? If the rams are 3 and 6 inches in diameter respectively, what 
are the lifting forces at the bottom and top ? AMiat are the fractional 
changes in the two cases ? If chains go from the cages vertically oA'er a 
pulley to a counter-weight, what ought to be the weight of the two chains 
per foot of their length ? 

Ans., 34-8 lbs. per square inch ; 3.043 lbs. ; 2,797 lbs. ; 3,687 lbs. ; 
•2J03 lbs. ; 8 per cent., 28 per cent., 3-00 lbs., 12-23 lbs. 

8. Show that if a tank has water in it to the same level, whether there 
are floating objects in it or not, its total weight is the same. If the weights 
of two tanks (with rams) are 160 and 144 tons, and they rest on rams a 
and B, each 3 feet diameter, what are the pressures at the bottoms of the 
rams r If the presses commimicate with one another, what is the difference 
of levels of the bottoms of the rams when there is balance ? Disconnect 
the presses; connect the press b with an accumidator whose ram is 21 
inches diameter, its end being now 30 feet below the end of b. What is 
tlie load needed for the accumulator if b is to be Kfted another 5 feet ? 

9. The areas of vertical cross-sections of the immersed part of a ship at 
intervals of 10 feet are a square feet, given in the follo^^TJig table. Find 
approximately the weight Wj of water displaced between every two neigh- 
bour sections. The weight of the portions of the ship bounded by the same 
sections are W2 ; what are the resultant loads Wo — w^ acting downward on 
the ship between, every two sections ? Draw a curve showing these values. 



A 


\1 

i 
1 




.30 


;'5: 

33,000 55,000 
37,000 [47,000 


100 


:100 

62,000 

54,000 


1100 

62.000 

57,000 




•jf> 




90 




78 


37,000 

39,000 


40 




Wl 
W2 


9.300 

23,000 


61,700 

,6.m 


59.000 

oS.rcO 


52,000 
47,00(r 


12,000 

»:'0oo 



10. Fifty rams, each of 14 inches diameter, begin to lift a gridiron 
from the bottom of a harbour, and so lift a ship which was floating 
directly above. The plane areas in square feet bounded by the water-line 
of the vessel, after the following numbers of feet of lift, are sho-^-n as the 



Lift in feet.! 

1 





1 


3 


6 


9 


12 


15 


18 


21 


24 


27 


80 


A 

P 


4,050 
250 


4,090 
283 


4,120 
350 


4,100 
450 


3,970 
547 


3,400 
637 


3,000 
717 


2,200 
780 


1,400 
824 


COO 
848 


200 
858 




800 



nim[ibers a. If the total apparent weight of the gridiron is taken to be 
constant at 860 tons, find the pressure in the presses for every position, 
and compare with the answers given in the table. 

173. Force due to Pressure of Fluids. 

Exercise 1. — Prove that Up, the pressure of a fluid, is constant, 
the resultant of all the pressure forces on the plane area a is Ap, 
and acts through the centre of the area. 

2. The pressure in a liquid at the depth h being tck, where 20 
is the weight of imit volume, what is the total force due to pressiu-e 
on any immersed plane area ? Let d e (Fig. 142) be the surface from 
which the depth h is measured, and where the pressure is 0. Let b q 



APPLIED MECHANICS. 215 

be an edge view of the area ; imagme its plane produced to cut the 
level surface of the liquid c in d. Let the angle e d c he called a ; 
let the distance d p he called x, and let n q, be called a? + Sa? ; and 
let the breadth of the area at right angles to the paper at p be 
called s. On the strip of area z . dx there is the pressure wh if h is 
r n, the depth of p, and h = x ^in.a; so that the pressure force on 
tlio strip is wx . sin. a . z . 5x, and the whole force is 



J DC 
X 
OB 



Y = w sm. a I X .z .dx . . . . (1). 

Also, if this resultant acts at a point in the area at a distance x 
from D, taking moments about n, 

i'nc 
X- . z . dx . . . . (2). 
BB 

fn c 
X . z . dx =■ AX, ii A is the whole area 
^ D B 
and X is the distance of its centre of gravity from d. Hence the 
average pressure over the area is the pressure at the centre of 
gravity of the area. ^^ ^ 

Observe in (2) that l x- . z . dx = i, the moment of inertia of 






the area about d. Letting i = k- a, where Jc is called the radius 
of gj'ration of the area about n, we see that 

F = tv sin. a . AX, p x = «^ sin. a . A k^. 

Hence x = k-jx. . . . (3), the distance from D at which the 
resultant force acts. 

Uxa>»ph.~Ji DB = and the area is rectangular, of constant 
breadth b, then 



-j: 



^ b 

X^ . dx = -X T) c^ 



and A = i . D c ; so that '^'-, 

/i;2 = 1 D 0^, also X =\t)C. 

Hence x = f do; that is, 

the resultant force acts at | of the way down the ■ 

rectangle from d to c, and the average pressure 

is the pressure at a point halfway down. 

It is an easily-remembered relation that we 
find in (3). For if we have a compound pendu- I 
lum whose radius of gyration is k, and if x is 
the distance from the point of support to its centre of gravity, 
and if x is the distance to its point of percussion, we have the very 
same equation (3). Again, if x is the length of the simple pen- 
dulum, which oscillates in exactly the same time as the com- 
pound one, we have again this same relation (3). These are 
merely mathematical helps to the memory, for the three physical 
phenomena have no other relation to one another than a mathe- 
matical one. 



216 



APPLIED MECHANICS. 



EXEECISES. 

1. Find the whole pressure of water upon a vertical dam 10 feet deep, 
30 feet wide. What would be the pressure of water of the same depth 
against a dam inclined at 45° ? Ans., 93,750 and 132,200 lbs. 

2. A water-tank is 13 feet square and 4 feet 6 inches deep. Find the 
pressure upon one of the sides when the tank is full. Ans., 8, 226-56 lbs. 

3. In the vertical plane side of a tank which holds water there is a 
rectangular plate whose depth is 1 foot and breadth 2 feet, the upper edge 
being horizontal, and 8 feet below the surface of the water. Find the 
pressure on the plate. Ans., 1,062-5 lbs. 

i. A rectangular tank, 5 feet square, is filled with water to a height of 
7| feet. A rectangular block of wood weighing 312-5 lbs., and having a 
cross-section of 5 square feet, is placed in the tank, where it floats in the 
water with this section horizontal. How much does the water rise in the 
tank, and what is the increase in pressure on a vertical side of the tank ? 

Ans., 2| inches, 489 lbs. 

5. A sluice-gate is 6 feet broad and 8 feet deep, and the water rises 3 
feet on one side above the lower edge, and 7 feet on the other side. Find 
the resultant pressm-e and centre of pressiu-e for each side of the gate. 
AMiat is the resultant of these and its position ? 

A?is., 1,682 lbs., 9,158 lbs. ; 2 ft. and 4f ft. below surface; 7,476 lbs., 
4-36 ft. from surface. 

6. A vertical wall, 10 feet high and 2 feet thick, weighing 153 lbs. per 
cubic foot, has to support the pressure of water (weighing 62-3 lbs. per 
cubic foot). How high may the water rise against one side of the wall 
without causing the resultant force at the base to pass outside the outer 

I? Aiis., 6-65 feet. 

7. A rectangular opening a b c d is made in the outer vertical face of a 
reservoir embankment and closed by a door, hinged along the lower 
horizontal edge a b. Find the pressure to be applied at the upper edge 
c D of the door, in order to shut against the pressure of the water. 
Breadth of door, 3 feet; depth, 4 feet; depth of upper edge belcw surface, 
8 feet. Ans., 3,489 lbs. 

8. In a hydraulic jack the ram is 5 inches, and the pump plunger | 
inch diameter; the leverage for working pump, 10 to 1; what is the 
velocity ratio? Experimentally we find that a force of 20 lbs. lifts 8,200 
lbs, ; what is the efficiency? A7is., 441|; 92^ per cent. 

9. Water at 750 lbs. per square inch acts on a piston 8 inches 
diameter : the velocity is multiplied by S by a four-sheaved pulley block ; 
what weight may be lifted by the chain if the efficiency of the mechanism 
is 55 per cent. ? In a lift of 40 feet, how many gallons of water are use J 'f 

Ans., 2,592 lbs. ; 10-87. 
174. Whirling Fluids.— If fluid 
rotates like a rigid body about the 
vertical axis o o (which is in the 
plane of the paper) with the angu- 
lar velocity a radians per second, at 
a point p'let there be a particle of 
weight IV lbs. ; let o r = .r feet. 

The centrifugal force is — arx in the 
Fig. 1^. ff 




APPLIED MECHANICS. 



217 



direction r d ; the weiglit is iv in the vertical direction p c ; and the 
resultant force is in the direction p e of the amount w Vl + a'^x'^jf/'^, 
the tangent of the angle d p e heing w -. a^w or g/ahv. 

Imagine many curves drawn such that their direction at any point 
reiiresents the direction of the resultant force there. Such curves are 




called lines of force. Let the point p be p fest above some datum 
level, and let as find the equation to the line of force which passes 
through p. The slope of the curve dy/dx is -tan. dpe (Fig. 143), 

or ^ = - -~- ; so that y = - - 
dx a^x ^ 



, _ ._„ ^ log. X -h e (1), where 

e is a constant. We see, then, that the lines of force are logarithm ic 
curves. If there is a curve to which p e is normal at the point p, 



218 APPLIED MECHANICS. 

its slope is positive, bein^ tan. a' p d. or ^ = — a; : so that the 
• = ^ dx g ' 

equation to the curve is y = —- x^ -\- c, where c is a constant, 

depending upon the datum level from which y is measured. This 
is a parahola, and if it revolves about the axis o o we have a 
paraboloid of revolution. Any sm-face which is everywhere at 
right angles to the force at every point is called a level surface, and 
we see that the level surfaces in this case are paraboloids of revolu- 
tion. These level surfaces are sometimes called equi-potential 
surfaces. It is easy to prove that the jpressure is constant every- 
where in such a surface, and that it is a surface of equal density ; 
■ so that if mercury, water, oils, and air are in a whii-ling vessel, their 
sm-faces of separation are paraboloids of revolution. 

The student ought to draw one of the Knes of force and cut out 
a template of it in thin zinc, o o being another edge. By sliding 
along o o he can draw many lines of force. Now let him cut out 
a template for one of the parabolas, and with it draw many level 
sm-faces. The two sets of curves cut each other everyrs-here 
orthogonally. Fig. 144 shows the sort of result obtainable where 
a a, bb, c c are the logarithmic lines of force, and a a, bb, c c are 
the level paraboloidal smiaces. 

175. The engineer seldom deals with other volumetric forces 
thtm those due to gravity and centrifugal force. By dealing in this 
way with centrifugal force he is able to 
A treat a rotational problem as a statical pro- 

, p. blem Whatever be the system of volumetric 

s. ' forces, we are supposed^ to know the lines 

of force in the fluid and a series of level 

or equi-potential surfaces cutting these 

' C lines at right angles. In Fig, 145, if ab c 

and a' b' g' be lines of force, and b b and 

c c be two level surfaces ; if f is the force 

on the fluid at o per unit volume and 

Fig. 145. B c = 5s ; if the pressui-e at b b' is ji; and 

at c c' it is ^ + 5/?, consider the equilibrium 

of the prism whose end of small area a is 

at B b' and other end of equal area is at c c' (we take the ends of 

equal area because we afterwards assume a and 5s to be smaller 

and smaller without limit). Evidently, as the volume is a . 5s, the 

volumetric force is f . a . 5s, and we have 

pa -^ ■£a.'8s= ip -\- ^p) a . . . . (1), 

Hence f . 5s = 5^?, or f = dp/ch .... (2). 

Example 1. — If gra%'ity alone acts and 5s is called 57i, so that A 
is measiu-ed vertically doicnicards fi'om a towards c, and we take 
F = w, the weight of a cubic foot of fluid, to be constant, being 

62-3 lbs. per cubic foot for water, then -jj = w . . . . {?>), or 

p = tvh + c where c is a constant. If we measure h from a level 
where we take the pressure to be /?„, then c = p^ and p — Pq = wh. 



^ 



APPLIED MECHANICS. 219 

Example 2. — Take w to be variable, say w = cp,Si rule that holds 

dp dp 

for gases at constant temperature, then (3) becomes -^ == c^, or — = 

c . dh, or log. j9 = c/i + where c is a constant. 

Let p = Po where h = o, then c = log. p^, and we have log. 
p/P(, = ch . . . . {4:). The actual value of c depends upon the 
constant temperature supposed to be maintained. 

Example 3. — Take tv to be variable, say w = cp^l"^ where 
y = 1-414 for air, being the ratio of the specific heats. This is 
the law which is much more likely to hold in a mass of gas than 
the constant temperature law. 

Then ^ = cp^y, OTp-^lf . dp = c . dh, or p'^ " l/V(l - Vy) = 

ch + c. It is easy to show that this leads to the result that the 
temperature increases in proportion to the increase of dej^th in the 
atmosphere. 

Example 4. — In our whu-ling fluid it is easy, since r = 

f7o 

W V I + a^xyc/- and - = V 1 + y7a%^ to find from (2) the law 
uX 

of i/, a one knows how to integrate. Take the simpler case, in 

which X is so great that the lines of force may be regarded as 

horizontal, and the level surfaces vertical circular cylinders. Then 

letting ds be called 5x, x being the radius of the path in which a 

particle revolves, 

p = —a^x, and ~a^x . dx dp. 

(1) If w\& constant, -^a?' i? + C where c is some constant. 
Let i?=i?o where x = x^, and we have p-2^o = ir- (^^- V) • • . • 

(2) If w = c2}'^, then — — z= ~\ p dp, or -^^-= —-^ — p, + c. 



-\f'- 



2g c ]' ^' 2ff [l-k) 

c (l-/v) a^{x^—x^) 
°'' "^ 2f " p'-^^-p^.^-K 

If we take k— — where 7 = 1*414 for air, and the easily found 

value of c for any given conditions, we can find the increase of 
pressure due to centrifugal force in a mass of whirling gas, as in 
the wheel of a fan when it is not delivering much fluid. 



220 



CHAPTER X. 

MACHINERY IX GENERAL. 

176. Me-chanism. — When the power of a steam-engine is dis- 
tributed through a factory, the distribution is performed bv 
means of shafts, spur and bevil wheels, belts and pulleys, and 
other kinds of gearing. As we are writing for men who have 
observed such transmission of energy, it is no part of our object 
to describe here what can be seen in any workshop. Perhaps 
no study is more useless from books alone than the study of 
mechanism ; whereas it is very useful and easy if we examine the 
actual thing, or make a skeleton model or a skeleton drawing. 

At the same time it is necessary to read books. The 
present book deals with the kinetics of mechanism ; but there is 
another part, a preliminary part, and students must read some 
book giving descriptions of contrivances and the mere kine- 
matics of the subject. AVe give a short sketch of certain 
important principles here, and in Chap. XXVI., because they 
are necessary for the proper understanding of our own division 
of the subject. 

177. Velocity Ratio. — In any machinery the velocity of a 
point may be calculated when the velocity of any other point 
is known. The number of revolutions per minute made by a 
shaft tells us the velocity of any point on any wheel or pulley 
fixed on the shaft ; the circumference of the circle described by 
such a pointy multiplied by the number of revolutions per 
minute, is evidently the distance moved through by the point 
in one minute. Now, when one shaft drives another by means 
of spur or bevil wheels, or by two pulleys and a strap, it is 
evident that the number of revolutions per minute made by one 
of the shafts, multiplied by the number of teeth of the wheel, 
or by the circumference or diameter of the wheel or pulley, is 
equal to the number of revolutions per minute made by the 
other shaft, multiplied by the number of teeth, or by the cir- 
cumference or diameter of the other wheel or pulley. This is 
evidently true, supposing that the strap does not slip on the 
pulley. Hence the rule — to find the speed of a shaft, driven 
from another by means of any number of wheels and pulleys, 
multij^ly the speed of the driving shaft hy the product of the 



APPLIED MECHANICS. 221 

diameters or nutnhers of teeth in all the driving wheels or 
pulleys^ and divide by the product of the diameters or numbers 
of teeth in all the driven wheels or pulleys. By the diameter of 
a spur wheel we mean the diameter of its pitch circle. Two 
spur wheels enter some distance into one another, and the 
circle on one which touches the circle on the other (the dia- 
meters of these circles being proportional to the numbers of 
teeth on the wheels), is called the pitch circle. The circum- 
ference of the pitch circle, divided by the number of teeth, 
gives the fitch of the teeth. 

178. Shapes of Wheel Teeth. — We know that if two spur 
wheels gear together, however badly their teeth are formed, so 
long as a tooth in one drives past the line of centres of a tooth 
in the other, their average speeds follow the above rule. But 
if we want the speed ratio at any instant to be the same as at 
any other instant, it is necessary to form the teeth in a certain 
way. Tlie curved sides of teeth ought to be cycloidal curves. 
The proof of this is not very difficult ; it is given in Art. 462. 
It is not usual to employ these cycloidal curves, for it is found 
that certain arcs of circles approximate very closely to the 
proper curves. The method of drawing rapidly the curved 
tooth of a wheel you will find taught by every teacher of 
mechanical drawing, you will find described in a great number 
of books, and you will see it in use in the workshop."^ You 
must remember that no study of books, and I may also say, no 
fitter's or turner's work that you may engage in, will make up 
for want of the experience which you would gain by actually 
drawing to scale a spur or bevil wheel, a bracket or pedestal 
with brasses, and a few other contrivances used in machinery. 
A worm and worm-wheel, that is, a screw, every revolution 
of which causes one tooth of a wheel to be driven forward, is 
sometimes used when we wish to drive a shaft with a very slow 
speed. If the worm-wheel has 30 teeth, it evidently makes 
one-thirtieth of the number of revolutions of the driving shaft. 

179. Skeleton Drawings. — When we consider the relative 
motions of, say, a piston and the crank which it drives, Ave 
come to something which it is not so easy to state without some 
little knowledge of mathematics. It is the same with all sorts 
of combinations of link work, and with. cams. Even a good 
knowledge of mathematics is only sufficient to give one a rough 
general idea of the relative motion in such cases ; and for the 

* Consult Professor Unwiu's "Machine Design " on the teeth of wheels. 



222 



APPLIED MECHANICS. 



0^^ 



study of any special case there is nothing so good as a skeleton 
drawing or a model. I give one example of the use of skeleton 
drawings— a crank and connecting- 
rod. Let A and b (Fig. 146) be the ends 
of a connecting-rod. As A moves 
from a-^ to c and back again, b de- 
scribes the complete circle, b^ d b^. 
Set off equal distances, b^ b^, b^ b, b b^, 
etc., and make b^ a^, b^ a^, etc., equal 
to the length of the connecting-rod. 
Then the points a-^ a^, etc., and 6^ 63, 
etc., show in a very good way the 
relative motions of A and B. When 
you have finished this exercise, work 
others in which, with the same length 
of crank, you have longer or shorter 
connecting-rods. You will get some 
such results as are shown in the upper 
part of the figure. In every case, if 
we imagine the crank to revolve uni- 
formly, the motion of A, the end of 
the connecting-rod, is show^n ; the 
distance from any one point to the next 
is passed over by A in the same in- 
terval of time. Simple Harmonic 
Motion (see Chap. XX Y.) is the name 
given to the motion of the piston- 
rod, when we imagine the connecting- 
rod to be infinitely long ; or rather, as 
we make the connecting-rod longer and 
longer, we get more and more nearly 
to this sort of motion. You see, then, 
that by skeleton drawings I mean 
drawings which show successive posi- 
tions of the diff"erent parts of a 
mechanism whose motions we want to 
study. You will find that an eccen- 
tric and its rod may be regarded as 
a crank (the length of a crank is the 
distance between the axis about which the eccentric is revolv- 
ing and its true cjentre), and a very long connecting-rod (the 



length of the connecting-rod bein^ 



the length of the eccentric 



APPLIED MECHANICS. 223 

rod measured to the true centre of tlie disc). The advantage 
derivable from skeleton drawings will be more obvious if you 
consider, in the above case of a crank and a connecting-rod, 
that A need not be the cross-head at the end of the piston- 
rod ; it may be the end of a lever, and so move in the arc of 
a circle ; it may be a slide moving in a slot of any curved 
form. One of the most instructive cases of skeleton drawing 
is a link-motion. Taking any good drawing of a link-motion 
to start with, find the relative motions of piston and of slide 
valve for various positions of the link. In the study of the 
motion of a slide valve it is much too usual to assume that the 
piston's motion is what is shown in Fig. 146 as simple harmonic 
motion. The reason of this lies in the ease with which it can 
be stated in mathematical language ; but it is incorrect. 

180. In this early part of our work we wish to confine our 
attention to energy and power calculation in the simplest mech- 
anisms ; the transmission of power by rotating shafts ; and the 
kinematic principles involved are very simple. In Chap. XXVI. 
we have something more to say about mechanism in general. 

EXEECISES. 

1. Two parallel shafts, -whose axes are to he as nearly as possihle 2 feet 
6 inches apart, are to he connected hy a pair of spur wheels, so that while 
the diiver runs at 100 revolutions per minute, the follower is required to 
run at 25 revolutions per minute. Find the diameters of the wheels and 
also the numher of teeth on each, if the pitch is 1 J inches. 

Ans., 48 inches, 12 inches ; 120 teeth, 30 teeth. 

2. A main shaft carrying a pulley of 15 inches diameter and running 
at 60 revolutions per minute communicates motion hy a helt to a pulley 
of 12 inches diameter, fixed to a countershaft. A second pulley on the 
countershaft of 8J inches diameter carries on the motion to a revolving- 
spiadle which is keyed to a pulley of H inches diameter. Find the 
numher of revolutions per minute made hy the last pulley. Ans., 150. 

3. On the crank-shaft of an engine there is a pulley 2 feet 6 inches ia 
diameter. By means of a helt this drives a pulley 25 inches in diameter on 
a second shaft, on which is another pulley, 24 inches in diameter, which 
drives another pulley, 15 inches in diameter, fixed on a third shaft. On 
this shaft is a pulley 25 inches in diameter, which drives one of 10 inches 
diameter on a fourth shaft. On this shaft is another pulley, 20 inches in 
diameter, which drives a pulley 8 inches in diameter fixed on a dynamo 
shaft. If the engine runs at 100 revolutions a minute, what will he the 
speed of the djTiamo ? Ans., 1,200 revs, per minute. 

4. In a screw-jack where a worm-wheel is used, the pitch of the screw 
is I -inch, the numher of teeth is 20, and the length of the lever which 
works the worm is 12 inches. What is the velocity ratio ? Ans., 2413-1. 

5. A wheel of 40 teeth is turned hy a winch handle 14 inches long, 
and gears with a rack having teeth of 1 inch pitch. If the axis of the 



224 APPLIED MECSANICS. 

wheel is fixed, what is the travel of the rack for two turns of the handle ? 

Ans., 80 inches. 

6. The tahle of a drilling-machine is raised by a worm-wheel in com- 
bination with a rack and pinion. The handle which worlcs the woi-m is 
12 inches long, the worm wheel has 30 teeth, and the pitch circle of the 
rack pinion is 4 inches in diameter. What is the lift of the table for one 
turn of the handle ? If the table and accessories w^eigh 500 lbs., what 
weight on the table would be balanced by a force of 12 lbs. api^lied at the 
handle, if 45 per cent, of the force applied be lost ? Ans., 42 inch 688 lbs. 

7. If the two wheels in the back-gear of a lathe have 63 teeth each, 
and the pinions 25 teeth, what is the reduction in the velocity ratio of the 
lathe sj^indle due to the back-gear ? Ans., 6 35 ; 1. 

8. The slide-rest of a screw- cutting lathe moves along the bed 14 
inches w^hile the loading screw makes 56 revolutions. What is the pitch of 
the screw thread? Afis., ^ inch. 

9. It is desired to cut a screw of f inch pitch in a lathe with a leading 
screw of 4 threads to the inch, using four wheels. If both screws be 
right-handed, what wheels would you employ ? 

10. It is required to cut a left-handed screw of 5 threads to the inch in 
a lathe fitted with a right-handed guide-screw of h inch pitch. Show 
how the change wheels might be arranged, and state the numbers of 
teeth on them. 

181. How a Shaft transmits Power. — I have refused to 
describe for you what you may see for yourselves at any 
time in workshops — how spur and bevil wheels and belts 
transmit power ; how there are arrangements for disengaging 
such gearing, and stepped cones for giving change of speed 
when belts are used ; how shafts are carried near walls 
or columns ; how machine tools work, and a hundred other 
matters about which a little obser- 
vation and drawing are of more im- 
portance than a large amount of read- 
ing. But there are some matters con- 
nected with machinery of great 
interest to you which you are not 
likely to observe unless I direct your 
attention to them. When a shaft 
transmits power it is in a state of 
strain ; it is in a twisted condition. 
The twist is not perceptible to the eye, of course, but methods 
have been arranged to show it to the eye, and measure it ; and 
it is found that the tioist in a shaft is jjroportional to the horse- 
poiver transmitted by the shaft divided by the number of revolu- 
tions per minute. Now to explain what I mean by a twist. Let 
a straight liiie be drawn along the shaft when power is not being 




APPLIED MECHANICS. 



225 



transmitted, then, if power be transmitted, the shaft will 
receive a twist, and this line will become a spiral line. The 
inclination, at any point, of the spiral line to its old position, 
is a measure of the twist. * When, instead of the ordinary 
coupling. Fig. 147, in which the two halves are connected by 
means of bolts, we use one, Fig. 148,t in which the two halves 
are connected by means of spiral 
springs, these springs get extended 
when the shaft transmits power. 
The yielding of the springs cannot 
be observed unless we make some 
arrangement like that shown, where 
the motion of a relatively to c 
causes the arm e to move and bring 
the bright bead b towards the axis. 
If everything is made dead black 
except the bead it will be seen de- 
scribing a circle of greater or smaller 
radius, and a scale with a sliding 
pointer enables us to measure ac- 
curately the distance moved inwards 
by the bead. The reading on the 
scale multiplied hy the number of 
revolutions of the shaft per minute 
tells us at once the horsepower actu- 
ally passing through the coupling. \ 

At Finsbury, the scale comes out from a wall, the shaft is 
about ten feet above the floor ; the observer stands on a ladder 
with a gas-jet behind him. It is quite easy to avoid error due 
to parallax, and to read with a very considerable amount of 




* The best measure of the twist is this angle of the spiral divided by the 
radius of the shaft, and the quotient is called the angle of twist. See Art. 294. 

t Ayrton and Perry's Dynamometer Coupling. 

X The total moment of the forces of the springs in pound-feet, or, as it has 
been called by Professor James Thomson, the torque, multiplied by the angular 
velocity in radians per minute, divided by 33,000, is the horse-power. Suppose 
that when one of the lengths of shafting is held fast we find the position of the 
bead when we hang weights on levers or round pulleys or wheels fastened to 
the other length ; a torque of 52*5 pound-feet will cause the bead to move 
radially inwards by a distance which we call '01 on our scale ; a torque of 105 
pound-feet causes the bead to move inwards a distance which we call "02 on 
pur scale, and so on. Such a coupling ought to be graduated by actual experi- 
ment. We generally have done it statically, and it is then necessary to 
eliminate friction by vibration, &c. In actual practice, friction is eliminated, 
because of the continual vibration of the parts. 



226 APPLIED MECHANICS. 

accuracy. It is exceedingly interesting to watch the bead when 
the loads are suddenly altered. 

It is a great pity that there should not be at least one such 
dynamometer coupling on every length of shafting in factories. 
We have it from a man who has made careful measurements 
that the loss of power in ordinary shafting is very great indeed. 
If the fact were continually before our eyes great improve- 
ments would certainly be effected. This is a function of 
measuring instruments (keeping defects prominently before us) 
which is very important. Mechanical engineers are largely in 
the habit of treating indicated horse-power of an engine as if it 
were the actual power given out by the engine. Steam-engine 
construction has improved enormously of late, mainly because 
electricians have been able to measure electrical power with 
great accuracy, and so the mechanical losses of power were 
brought prominently into notice. 

182. Belts. — If the pulley b, Fig. 149, is driven from A by 
means of a belt, you must remember that there is a pull in the 

part of the belt m, as well as 
in the part n. These two 
pulls are generally pretty 
great, as you know, but if 
you could measure them ac- 
curately you would find that 
there is more pull in n, else 
Fig. 149. A would not turn. It is the 

difference of these pulls which 
concerns us. You may, perhaps, understand this better from 
Fig. 150. The pull in a m is the weight of M, say, 20 lbs. The 
pull in a n is the weight of N, say, 50 lbs. If n falls two feet, 
M rises two feet, and the work done upon the pulley and which 
it transmits through the shaft somewhere else is 50 x 2, or 100 
foot-pounds, minus 20 x 2, or 40, the difference being 60 foot- 
pounds. In fact, it is the difference of pull in the two cords, 
30 lbs., multiplied by the space passed over by the cord, 2 feet; 
result, 60 foot-pounds. 

The horse-power given by a belt to a indley is, then, the 
difference of pull in the belt on the two sides of the pulley ^ 
multiplied by the speed of the belt in feet per minute, divided 
by 33,000. 

This is only a particula.r case of the general rule. If m is 
the sum of the moments of a number of forces tending to cause 




APPLIED MECHANICS. 



227 




rotation, about the axis, in loound-feet, and if Sl is the angular 
velocity in radians loer r)iinute, then Ma is the work in foot- 
pounds per minute, so that Ma -^ 33,000 = H, the horse- 
power .... (1). 

Example. — In our dynamometer coupling, if there are four 
springs, each exerting a force of 160 lbs. ; the distance of the 
axis of each spring from the axis of the shaft being 0*7 foot, 
the turning moment is 4 x 160 x 0-7, or 448 pound-feet. If 
the shaft makes 150 revolutions per minute, this means 
150 X 27r, or 942 radians per minute; and hence, 448 x 942 
~ 33,000 = 12-8 horse-power. 

Example.- — An ordinary flange coupling has six bolts, at 
07 foot from the axis; what force is resisted 
by each bolt (it tends to break by shearing, 
Art. 281), when 60 horse-power is being trans- 
mitted at 120 revolutions per minute? Answer. 
— If F is the force, 6 F x 0*7 is the moment 
in pound-feet; this, multiplied by 120 x 27r 
~ 33,000, is equal to 60; and hence, f = 33,000 
x 60 ^ (6 X 0-7 X 120 x 27r), or Fr=625 lbs. 

When we know the maximum horse-power 
at the minimum speed (observe that torque or 
turning moment depends upon power -^ speed), 
we can calculate the maximum f for each bolt, 
and our knowledge of the strengtli of materials 
(Art 284) enables us to say if the bolt is 
strong enough. 

The general rule appears in many special m^ 
forms, of which we now have one example in 
belting ; and just as it enables us to calculate 
the strength of bolts in a flange coupling, or 
the size of the shaft itself (Art. 296), or tells 
us how to order the springs of a dynamometer 
coupling, so it here tells us how to find the proper size of 
a belt. We have here (n— m) v -^ 33,000 = h, the horse- 
power. The greatest pull in the belt is n lbs., and it is this 
which determines how strong the belt must be. Hence, we 
must answer the question : — 

If it is the difference of pull that produces turning, why is 
there so great a pull even in m. Fig. 149^ as we usually findl 
Refer again to Fig. 150. If we want the difference between m 
and N to be 30 lbs., why not make m have no weight at alL ^nd 




Fig. 150. 



228 



APPLIED MKCHANICS. 



N may then be only 30 lbs. ? Evidently we should not be able 
to get friction enough, and the weight N would fall, causing 
the cord to slide on the pulley ; in fact, the friction between 
the cord and pulley must be more than 30 lbs., else there will 
be slipping; and to produce this friction it is necessary to have 
weight at M as well as at N. If we allowed the cord to lap 

round more of the 
pulley, the necessary 
friction might be pro- 
duced with a less 
weight at m. To get 
an idea of the friction 
between a cord and a 
pulley, arrange a pulley, 
or other round object, 
p, as in Fig. 151. Fix 
it firmly. Place a 
weight at m, say 1 lb. 
Now place weights in 
the scale-pan at N, until 
the cord just slips 
slowly. Say we find 
3 lbs. to be necessary. 
The difference between 
N and M, or 2 lbs., is the 
friction. Now put twice 
the f or] ner weight at m; 
you will fi.nd that about 
twice the former n will just cause slipping, so that the friction 
is doubled. In fact, we have our old law, " friction is propor- 
tioned to load." But now let us see how friction depends on 
the amount of lapping of the cord. In your first experiment 
measure the cord actually in contact with the post P. Suppose 
it to be 4 inches : now, keeping M, 1 lb., let the cord lap round 
more of the post P, say 8 inches this time, and. find the weight, 
N, which will just produce a slow sliding. You will find it 
to be 9 lbs. If the cord touches on 12 inches of the post 
p, you will find that 27 lbs, at n will be necessary to slowly 
overcome the friction. It is only by actually trying this 
experiment for yourself that you will get a clear idea of how 
rapidly the friction increases with the amount of lapping. It 
is on this account that Cae man can check the motion of the 




151. 



APPLIED MECHANICS. 



229 



largest vessel by simply coiling a rope a few times round 
a post. 

The apparatus, Fig. 151, is so arranged that any required 
amount of lapping may be given to the cord round the fixed 
post p. In an actual experiment, the fixed weight m was 50 
grammes. By means of the pulleys the amount of lapping 
round p was varied, and weights were placed in n, in each case 
just sufficient to overcome the friction and raise m slowly, as 
above described. The following are the results of the whole 
series of experiments : — 



Number of times the cord 
laps round. 


Tlie weight required at n to 

overcome friction and the 

weight of M. 


Logarithms of the ratio of N 
to M. 


^ 


80 


0-2041 


1 


105 


0-3222 


1 


150 


0-4771 


H 


200 


0-6021 


H 


255 


0-7076 


i| 


330 


0-8195 


2 


'400 


0-9031 


H 


500 


1-0000 


2| 


700 


1-1461 


2| 


1,000 


1-3010 


3 


1,150 


1-3617 


H 


1,500 


1-4771 



Plotting the first and third columns on squared paper, we 
find that a straight line passes nearly through all the points. 
From this line we deduce the equation — 

^-2-2 log-, 

M 

where n is the number of times the cord laps round. From 
this it is easy to show that the coefficient of friction, ju, between 
the cord and the pulley is -166. 

You must then remember that the tension in m. Fig. 149, is 
necessary to produce as much friction as will prevent slijDping. 
If ever the excess pull in n is greater than the friction, there 
will be slipping. If the belt slips, there is energy wasted, 
which you can calculate if you know the force of friction, and 
multiply by the distance through which slipping occurs. 



230 



APPLIED MECHANICS. 



183. The law is this. If fi is the coefficient of friction hetween 
the cord or belt and the pulley ; if Hs the length of the cord or belt 
which touches the pulley, say in inches ; and r the radius of the 
pulley in inches, then , 

N and M being the pidls in the belt or cord on the two sides of the 
pulley. Ijr is the angle of lapping stated in radians. 

This rule is arrived at mathematically in the following way. 

Let A p Q B be the part 
GL of a pulley touched 

by the belt ma p o, b n. 
Imagine the pulley 
fixed, and slipping 
occurring because the 
tension at n is greater 
than at m. Let the 
angle a o p be called d. 
Let the tension in the 
belt at p be t. We 
study what occurs at 
the place p a, and we 
greatly magnify p q 
in Fig. 153. What 
are the forces acting 
upon the short piece of belt p q ? We have a pull t + S t at q, 
and a pull t at p, and forces acting radially from the pulley, 
their resultant being x ; we have also friction whose amount 
is ;it X if /i is the coefficient of friction, and this friction is what we 





T^n 



Fig. 153. 



Pig. 154. 



overcome by the excess tension 5t. To find x, assume St = o, 
and no friction (it will be found that these tenns become less and 
less important as the distance p a is made less and less). Let the 
angle a o a be called 6 + 50, so that p o q = S 9. The three forces 
T, - X, T being in equilibrium, let them be represented in direction 
and sense by the sides of the triangle shown in Fig. 154, and it is 
evident that x = t . S0. Hence the force of friction is ^i . t . 50, 
which is overcome by 5t. When slipping is just occurring, 
;u . T . 50 5t . . . . (1) ; 

or rather, because the theory is true only when 50 is thought to be 
smaller and smaller without limit, 

This is an example of the compoimd interest law. " The rate of 



APPLIED MECHANICS. 231 

increase of t as increases is proportional to t itself." In any- 
elementary book on tlie calculus it is stown tliat (2) is the same as 
log. T = fid -\- constant. 
Putting T = M wlien 6 = 0, and t = n when = a o b, we 
find 

log. N — log. M = /* . A B . . . . (3). 

This is the rule given above. Assuming that there is a constant fi, 
and that we know its value, this rule enables us to design belts and 
ropes to be strong enough to transmit a given amount of horse- 
power H at the belt velocity v feet per minute. For we have 
already used (n - m) v -t- 33,000 = h . . . . (4). If the angle 
A o B be ca.lled d for shortness, it is easy to deduce from (3) and (4) 

«=?M^.M<./(„.._1)....(4). 

This is the tension in the belt where it is greatest, and it is from 
this value that we calculate the strength of the belt. Thus if b 
and f are the breadth and thickness of the belt in inches, and if / 
is the safe load per square inch of section for the material, we 
make btf equal to the calculated n. We take /about 330 lbs. per 
square inch in leather belting, and this leads to the easily 
remembered rule, gi"^^ng H in square inches, 

bt = ^-^ eM6>/(e/^0 - 1) (5). 

184. In a great number of cases the least angle of lapping 
when one shaft drives another by belting is -|ths of the circum- 
ference, and ju may be taken as 0'3 ; and so we are led to the 
common, easily remembered shop rule, bt = 200 H -f- v . . (6). 
It is to he remembered that b is the breadth, t the thickness of 
the belt in inches ; h is the horse-power transmitted, v the 
velocity of the belt in feet per minute ; fj. the coefficient of 
friction, and d the angle of lapping in radians. 

Leather belting" is seldom more than 4. feet wide, even in 
America, where the widest belting is employed. A single 
leather belt is about a quarter of an inch thick, and the 
material will stand a pull of from 700 to 1,200 lbs. per inch of 
its width, before breaking. A single belt at an ordinary laced 
joint may be taken to stand a greatest working pull of about 
one-third of the strength of the leather. An ordinary laced 
joint, with splice, has about twice the strength of an ordinary 
grip-fastened joint, and not quite twice that of a butt laced 
joint. Many engineers take 80 to 90 lbs. per inch of width of 
a single belt as the usual pull on the tight side. 

Strips from the best parts of the tanned hide are cemented 
at long splices to form belts : vulcanised indiarubber alone, or 
with intermingled canvas cotton, and various other kinds of 



232 APPLIED MECHANICS. 

waterproof belting are used. There is always slipping or 
creeping, so that it is not quite accurate to take the velocity 
of the belt as being equal to tliat of the pulley. (See Art. 70.) 
Belts are in use which are made of a greab number of little 
links of thick leather on wire pins, forming a chain. Hemp and 
cotton ropes are now often used, instead of belting. Many Y 
grooves in the rim of a drum receive each its own rope, which 
is usually lying wedged in the groove, with not much tension 
when there is no motion, except what is due to its own weight, 
the span being 20 to 60 feet. As in belts conveying a con- 
siderable amount of power, the tight, or n side, is underneath, 
as this gives more lapping. The strengths of ropes and 
chains are given in Art. 266, but in rope-driving our ability 
necessitates some such rule as (see Art. 183) 
N — M = 8 ^2^ 

when g is the girth of the rope in inches, and n and m are in 
pounds. The wedging of the ropes in grooves causes the 
coefficient of friction to be virtually of its usual value divided 
by the sine of half the angle of the groove, and as this is 
usually 45°, we take fi to be about 0*6, and sometimes more. 
Splices are about 10 feet long. The speed is usually about 
4,000 feet per minute. The bending and unbending of the 
rope is what seems to shorten its life, which is usually one of a 
few years. There is probably considerably more waste of 
energy by friction than when toothed gearing is used. 

With wire ropes (strands of wire round a hempen core) the 
wedging in grooves is much more hurtful, and hence reliance 
is placed upon greater velocities, even to 6,000 feet per minute. 
Under usual conditions the life of a rope is one of less than a 
year. In calculations of the tensions due to the weights of 
ropes, we may take it that the hanging curve is practically 
parabolic (see Art. 133). The bottoms of the grooves on the 
pulleys are usually of leather, wood, or gutta-percha, and jj. 
may then be taken as 0-25. Each relay being about 1,000 
yards, we may take its efficiency as about 95 per cent. 

Exercise. — If there are nine relays, what is the total 
efficiency'? Answer. — 0*95 raised to the 9th power, or 0*63. 

At Oberursel a rope is used whose diameter is 0-6 inch, of 
36 wires each 0-06 in diameter; pulleys, 12 feet in diameter, 
placed 400 feet apart; 94 horse-power is transmitted 3,000 
feet, the rope travelling at 4,400 feet per minute. 

185. Chain Gearing. — The use of tricycles and bicycles has 



APPLIED MECHANICS. 233 

caused a veiy great development in this non-slipping, certain, 
and efficient method of transmitting power. It is likely to 
have a very large further development for large powers in 
factories on account of its success for small power, not merely 
in cycles, but in electric -motor, tramcar and locomotive work. 
No tension is needed on the slack side. 
EXEECISES. 

1. A pulley of 3 feet diameter receives 10 horse-power at 150 revolu- 
tions per minute ; find the difference of pull between n and m. If the 
lapping is 0*4 of a circumference, and /i = 0-25, find n/m. Now find n. 

Ans., 233-4 lbs. ; n=501-3 lbs. 

2. Use the above workshop rule to find the breadth of a single belt 
(^ = 0-4 inch, say) to transmit 20 horse-power at a velocity of 1,300 feet 
per minute. ^ns., 77 inches. 

3. A rope is wrapped three times round a post, and a Tveight of 12 lbs, 
is hung from one end. Find the least pull applied at the other end neces- 
sary to raise this weight U. fx = '3. What weight would be required to 
just prevent the 12 lbs, from slipping down? Ans., 3,440 lbs. ; 0*042 lb. 

4. The pulley on an engine shaft is 5 feet in diameter, and it makes 
100 revolutions per minute. The motion is transmitted from this pulley 
to the main shaft by a belt running on a pulley, and the difference in 
tension between the tight and slack sides of the belt is 115 lbs. What is 
the work done per minute in overcoming the resistance to motion of '^he 
main shaft ? What is the horse-power transmitted ? 

Ans., 180,714 ft. lbs. ; 5-47. 

5. 50 horse-power is being transmitted by a belt moving at a speed of 
70 feet per second. What width of belt will be required if its thickness is 
0-6 inch, assuming the maximum working stress to be 330 lbs, per square 
inch, and the tension on the tight side being double that on the slack side ? 

Ans., 4 laches. 

6. A pulley 3 feet 6 inches in diameter, and making 150 revolutions a 
minute, diives, by means of a belt, a machine which absorbs 7 horse- 
power. What must be the width of the belt so that its gxeatest tension 
shall be 70 lbs. per inch of width, it being assumed that the tension in the 
driving side is twice that on the slack side ? Ans., 4 inches. 

7. A rope pulley, carrying 20 ropes, is 16 feet in diameter, and trans- 
mits 600 horse-power when running at 90 revolutions per minute. 
Taking jx = 0*7 and the angle of contact 180°, find the tensions on the 
tight and slack sides of the rope. Ans., 246 lbs. ; 27"2 lbs. 

186. Transmission and Absorption Dynamometers. — I 
have already described to you an instrument which allows us 
to measure the horse-power transmitted by a shaft. I am in 
the habit of employing a somewhat similar arrangement for 
measuring the power transmitted by a belt to any machine. Ifc 
is shown in Tig. 155, and is easily understood from the descrip- 
tion of Fig. 148. I can take it near any machine, and drive the 
machine through it, using two belts instead of one. G is a 
loose pulley. A belt drives H, which drives the plate e through 



23 i 



APPLIED :\IECHANICS. 



four spiral springs B. The plate e is keyed to a shaft carried 
on the frames c and d, and the pulley r is keyed on the shaft. 
A belt from f, therefore, will drive any machine. When much 
torque is acting, the springs b become extended, causing a 




relative motion of e and H, and this motion is shown by the 
bright bead a, at the end of the lever i a, approaching the 
axis of rotation. A fixed scale attached to the frame c 
allows the motion of a to be measured. 

In Fig. 159, shoAving the Froude or Thorneycroft Dynamo- 
meter, the pulley d drives the pulley f by a belt which passes 



APPLIED MECHANICS. 



235 




also round the two pulleys A and b. These pulleys are on a 
frame A B L, which is pivoted at e. The tension n, on the tight 
side, is greater than m, and a measurable force p must be ex- 
erted to keep the frame in the position 
shown. Evidently 2i!f. a = 2m. a -\- P.b 

P.6 
so that N — M=7i — • 
2a 

This transmission dynamometer is 
specially useful when small powers are 
to be measured in the laboratory. In 
the Hefner-Alteneck form, the long 
and nearly horizontal belt driving a pulley passes round two 
guide pulleys d and c on a suspended and balanced frame. 
When the angle made by the slack parts mm is the same 
as that made by the tight parts n n, a measurable force f is 
required to maintain the frame in its symmetrical position. 

Exercise. — If the angle between m and m is 180 — 0, and 
there is the same angle between N" and N, and the pull F is 
symmetrical, show that F = (n - m) if is small. In truth, 
F = (n — m) 2 sin J d. 

187. Absorption Dynamometers are used to measure the 
power given out by steam-, 
or gas-, or oil-engines, or 
electric or other motors. 
They measure the power, 
consuming it as they do so. 
One of Professor James 
Thomson's forms, as ar 
ranged for measuring the 
power given out by a small 
electro - motor, is shown 
in Fig. 157. The motor 
drives the grooved pulley 
A, and the pulley B turns 
along with A. A cord hangs 
lapping round part of B, 
and carries at its one end 
a scale-pan m, containing a 
weight. The other end n' 
is pulled by means of a 

piece of metal fastened to the rim of a loose pulley c, which 
has a weight N always acting upon it, tending to turn it round. 




Fig. 157. 



236 



APPLIED MECHANICS. 



Evidently the cord is pulled with a weight 3i at one end, and 
a weight x at the other. If now there is slipping between the 
cord and b, the friction is measured by the diflereuce of the 
weights y and M. If m is 100 lbs. and n is -iOO lbs. the 
friction is 300 lbs. If the pulley has a circumference of 
2 feet, and makes SO turns per minute, the amount of slipping 
is 80x2, or 160 feet per minute, and the work done against 
friction is 160 x 300, or 48,000 foot-pounds per minute — 
that is, 1*45 horse-power. In this case all the poicer is wasted 
in frictioji, and this is called an Absorption Dynamometer 
because it measures the power but absorbs it in doing so ; 
whereas the coupling of Fig. 148 and the dynamometer of 
Fig. 155 are called Transmission Dynamometers, because they 
measure the power transmitted through them whilst working 
any machines. 

Should we try to use merely a cord or belt with two 
diflerent weights x and M at its ends (as in Fig. 158) as an 
absorption dynamometer, the cord or belt slipping on the 
pulley, we should find that, after adjusting the ^'eights, 
when we leave them hanging on, x will gradually overcome 
M till it touches the floor, and after that we are ignorant 
of y — M. In fact, the co-efficient of friction gi-adually altei-s. 
In Fig. 157 this is automatically 
counteracted by a change in the 
amount of lapping. Sometimes, 
instead of the lighter weight ii, we 
have used a spring balance and a 
dash-pot to still the jerky motion 
which occurs, but in this case we 
must take readings of the changing 
values of M: and again, it is usually 
very important during a test to keep 
y— 51 constant. We have met with 
quite wonderful success by adopting 
the following very simple expedient : 
— A B is a grooved pulley, and a cord 
with scale-pans hangs round it as in 
Fig. 158. At p there is a knot in 
the cord, or an excrescence of any 
kind which, when it is drawn up 
bv the crradual falling of x, will just slightly jam itself 




Fig. 15S. 



the 



When measuring the power from a thi-e©- 



APPLIED MECHANICS. 



237 



quarter horse electric motor we have kept the same weights m 
and N in the scale-pans for an hour at a time, a slow gentle rise 
and fall taking place now and again. Before we discovered 
the value of the knot we went to great 
trouble in having a thick rough cord from 
p to Q spliced to a thinner one, so that 
when N fell a greater lapping of the 
rougher cord made up for a lessening of 
the friction on the smoother one, but as a 
matter of fact the knot is excellent. We 
have used the same expedient with a can- 
vas belt on a large drum when measuring 
26 horse-power ; here a leather 
lace passed carelessly a number 
of times through the belt in- 
troduced sufficient resistance 
at p to prevent n falHng. 

In all these cases soapy 
water is kept trickling over 
the rubbing surfaces to carry 
off the heat, for all the 
mechanical energy which we 
measure is converted into 
heat. The water ought to 
be soapy, because when pure 
water acts as a lubricant 
there is considerable, and 
seemingly spasmodic, varia- 
tion in the friction. A brake 
on the principle of n, a 
weight, and m, the pull of a 
spring balance, is sometimes 
used even for as much as 
50 horse-power, N being at- 
tached to two ropes going 
about three-quarters of the 
way round an ordinary 
pulley, and kept apart on the 
smooth rim by distance pieces 
of wood which fit the edges 
of the rim, one rope com- 
pleting the round so as to 




r' 



238 



APPLIED ]Mi:ClJA\ICS. 



come between the first two on leaving, m is then an upward 
pull, and N downward. 

Fig. 160 shows a brake which maybe used for large powers, 
and we have used it with satisfactory results. It consists of 
blocks of wood on hoop iron, forming a brake strap, the two 
ends of which D and e are fastened to the lever d h, a small 
angular change in the position of which produces great increase 
in tightness. A. load w lb. is hung on at c and a spring balance 
acts at H. The pull in the spring balance, w lb., is usually 
small. It is evident that w.R — z(;ris a moment in pound 
feet which, multiplied by the angular velocity in radians 
per minute, and divided by 33,000, gives the horse-power. 
Lastly, Froude's liquid dynamometer may be employed up to 

very high powers ; 
the heat is devel- 
oped in a consid- 
erable mass of 
water which 
keeps cool. For 
the latest and 
best form of this, 
students are re- 
ferred to a paper 
by Prof. Reynolds 
in the Transac- 
tions of the Royal 
Soc, May, 1897. 
If any student has trouble in seeing why we assume a 
virtual surface of rubbing coinciding with the centre of the rope 
in Fig. 158, let him return to the consideration of Fig. 150 
Here it is obviously the velocity of the centre of the rope that 
is taken, and we calculate this from angular velocity and virtual 
diameter of the pulley. ISTow suppose the pulley of Fig. 150 
to be at rest and the rope slipping ; the 
same amount of power is now wasted 
that used to be given to the shaft; there 
will be exactly the same forces of fric- 
tion and waste through friction if the 
pulley moves and the weights do not. 

Generally, let a b be a revolving 
wheel with centre o, and let forces f, 
w and w keep the brake block p Q r from moving with the 




Fig. 160. 




APPLIED MECHANICS. 239 

wheel. Let there be no other forces acting on the brake 
block which exercises moment about o. 

How much power is being wasted at the rubbing surfaces 1 
It is the sum of all the forces of friction multiplied by their 
respective rubbing distances (we need not here speak of pounds, 
feet, etc., or the unit in which power is measured). But this 
is the same as the sum of all the moments about o of all the 
forces of friction multiplied by the angular velocity, and the 
sum of the moments of the forces of friction is measured by 
w. p + F. R - ^y. OQ., because the block is in equilibrium 
under the action of the forces of friction and w, F and 7v (see 
Art. 98). 

188. Example. — In an Ayrton-Perry Coupling, the bright 
bead does not begin to move inward till a certain horse- 
power is reached. If a radial motion of 10 inches corre- 
sponds to a lengthening O'l inch of each spring; if 
there are eight springs, whose axes are 10 inches from 
the centre, and the shaft runs at 200 revolutions per 
minute ; if the maximum horse-power is 250, and if it is to 
alter from 200 to 250 for an 18 inch motion of the bead, how 
ought the springs to be ordered? If f is the pull in each 
spring at H horse-power, 8xFxlO-r-12in pound feet, multi- 
plied by 200 X 2 TT -^ 33,000 gives h, or f = 3-9i h. Hence the 
range of pull in each spring is from 3-94 x 250, or 985 lbs., to 
3*94 X 200, or 788 lbs., and for this the bead has a range of 18 

inches; and therefore each spring will yield by 18 x — , or 

0*18 inch. Thus the spring yields 0*18 inch for a difference 
of pull of 985-788, or 197 lbs., and will therefore yield 
985x0-18-^197, or 0-9 inch for the maximum pulL We 
therefore order eight springs, each of which has a greatest work- 
ing pull of 985 lbs., and is then to have an axial lengthening 
0*9 inch. It will also be necessary to tell the spring-maker 
what the exact length of the spring is to })e between the end 
pins, and possibly its outside diameter. It is usual to arrange 
adjustable stops to keep the springs stretched, so that they 
shall not elongate for any force less than 788 lbs. When the 
springs are long we use diametral wires to prevent any action 
due to centrifugal force, and in some cases of very fluctuating 
loads we have used dash-pots. 

Example. — The middle of a key is 2 inches from the 
centre of a shaft ; the key is the only fastening of a wheel 



240 APPLIED MECHANICS. 

which makes 100 revolutions per minute, and gives out 10 
horse-power. What is the force transmitted by the key^ 
Answer. — The force is f if f x 2-^12 (the moment in pound 
feet), multiplied by 100 x 2 tt -^ 33,000 == 10. That is, 
F ^ 3,016 lbs. 

Uxar/iple. — A 6-feet drum at 150 revolutions per minute 
drives through a Hefner Alteneck Eynamometer. A spring 
balance registers f as o when no power is being transmitted, 
and registers it as 60 lbs. when we know that 25 horse-power is 
being transmitted. Find if (n — m) = f. We know that the 
velocity of the belt is 6 tt x 150, or 2,827 feet per minute ; so 
that N - M is 25 X 33,000 ~ 2,827, or 292 lbs. Hence, d is 
60 -f- 292, or 0*206 radians, or 11*8 degrees. 

Example. — A dynamo machine going at 420 revolutions 
per minute is supported on knife edges. The dri\dng-belt 
causes the machine to tilt, and the tilting tendency is prevented 
by a weight w of 15 lbs. suspended from an arm 5 feet from 
the axis, measured horizontally. Find the horse-power given to 
the machine. The machine gives out 39-2 amperes at 98 volts. 
What is its efficiency? Answer. — The turning moment is 
15 X 5 pound feet, and therefore the horse-power is 15 x 5 x 
400 X 2 7r -f 33,000, or 5-71. The electrical horse-power 
given out is 39*2 x 98 -f 746 = 5-15, The efficiency is there- 
fore 5-15 -:- 5-71, or 0-902, or 90-2 per cent. 

Exarrfple. — A brake block is kept at rest by certain forces, 
500 lbs. acting at 2 feet from the axis, 30 lbs. acting at 8 feet 
from the axis, 50 lbs. acting at — 5 feet from the axis (we mean 
here that the moment of this last force is opposite to the 
others). The shaft makes 120 revolutions per minute. Find 
the horse-power consumed. The moment is 500 x 2 -f- 30 x 8 
— 50 X 5, or 990 pound-feet. This, multiplied by 120 x 2 tt 
-f- 33,000 gives 22*6 horse-power. 

Example. — The following readings were taken when testing 
an electric motor. Find the electric power supplied and the 
mechanical power given out in each case. The knot dynamo- 
meter was used ; each string was found to be almost exactly 
6-1 inches horizontally from the centre. If n is the number 

of revolutions per minute (n — m)x-— x27rxn-=- 33,000 

= H, or (n-m) n -h 10,000 = h nearly. The student will 
remember that 1 ampere X 1 volt = 1 watt, and 746 watts = 1 
horse-power. 



APPLIED MECHANICS. 



241 













Electrical 


Brake 




Amperes. 


Volts. 


N 


M 


n 


Horse- 
power. 


Horse- 
power. 


Eflaciency. 


8-4 


28-1 


4 


1-6 


1,504 


•318 


•162 


•51 


7-5 


30-4 


4 


2-4 


1,796 


•304 


•129 


•42 


13-1 


19-8 


18 


13 


572 


•349 


•128 


•37 


16-4 


31-0 


19 


12 


1,008 


•681 


•316 


•46 


15-3 


30-7 


17 


10 


1,048 


•632 


•329 


•52 



EXEECISES ON BEAKES. 

1. In an AjTton and Perry knot-trake (Fig. 158) the centres of the 
cords to N and m lie respectively 6 and 6^2 inches horizontally from o (it 
is usual to take these distances equal and equal to the radius of the circle 
of the centre line of the cord on the pulley, but very careful measurement 
will detect slight differences due to stiffness of cord and effect of knot) . 
The weights are n = 30 lbs., m = 5 lbs. The pulley makes 400 revolu- 
tions per minute; find the horse-power. Here we had better take 
moments, as the distances are unequal. The moment is (30 x 6 -i- 12) — (5 
+ 6 2 -~ 12), or 12-42 pound-feet. The speed is 400 x 2 tt, or 2,514 radians 
per minute, and 12-42 x 2,514 -h 33,000 = 0-946 horse-power. 

2. In a rope-brake on a flywheel of 8 feet diameter, the ropes being 1 
inch thick (so that the moment due to friction is as if the rope had no 
thickness, but the wheel is 8*083 feet in diameter), the load is 500 lbs. 
The pull in the sjjring balance varies from 10 to 20 lbs. during the test. 
The wheel makes 105 revolutions per minute ; find the horse-power. The 
velocity of the virtual wheel is 8-083 tt x 105 feet per minute, and this, 
multiplied by 490 or 480, divided by 33,000, gives a power varying from 
39-6 to 38-8 horse-power. 

3. The power of an engine is tested by passing a belt over the fly-wheel, 
which is 5 feet in diameter. One end of the belt is secured to a spring 
balance, and a weight of 300 lbs. hangs on the other end. What is the 
brake horse-power when the balance registers 180 lbs. and the fly-wheel 
makes 150 revolutions per minute ? A^is., 8|-. 

4. In Exercise 3 the pull in the spring balance gradually alters during 
the test to 200 lbs. This is due to a change in fi, the co-efiicient of 
friction. Find the two values of /i. Ans., 0-159 ; 0-129. 

If the extreme tensions 300 lbs. and 180 lbs. were maintained con^ 
stant, as in the James Thomson dynamometer (Fig. 157), and the lapping 
was 180° to begin with, what would it be at the end ? Ans., 222°. 

5. The diameter of a steam cylinder is 8 inches, the length of crank 
9 inches, the number of revolutions per minute is 150, and the mean 
effective pressure of the steam is 35 lbs. Find the indicated horse-power. 

The same engine is tested by a brake (like Fig. 160) on the crank 
shaft, R being 2^ feet, w being 310 lbs., iv being 15 lbs., and r being 
4^ feet. Find the brake horse-power and the working efficiency of the 
engine. Ans., 24 ; 20-2 ; 0.84. 

6. Find the horse-power given out by a steam-engine driving a 
Froude dynamometer which discharges 700 gallons of water per hour 
with a rise of temperature of 18*5 Centigrade degrees. Am., 92. 



^^4.--' 



CHAPTER XI. 

KINETIC ENERGY. 



189. We sometimes assume that our readers know quite 
vvell the fundamental principles of mechanics, and then, again, 
we assume that they do not. We hope that they agree with 
us that we are right in proceeding in this way. 

When a weight, A Fig. 22, in falling lifts a weight b by 
the use of a machine inside the box c, let us consider the store 
of energy at any instant. The store of (energy consists in — 
first, the -potential energy of A — that is, the weight A in pounds, 
multiplied by the distance in feet through which it is possible 
to let it fall to some datum level ; second, the potential energy 
of B, which is the weight of B multiplied by the distance through 
which it is possible to let b fall ; third, the energy of motion, or 
kinetic energy, of everything which is moving — namely, a, b, 
and the parts of the mechanism. We are supposing that there 
are no other weights which can fall or rise, and that there are 
no coiled springs or other stores of energy in the jnechanism. 
Now, if A is just heavy enough to maintain a steady motion, 
the kinetic energy remains the same ; so that, whatever energy 
is given out by a in falling is in part being given as potential 
energy to b, and is in part being wasted in friction. But 
suppose A to be heavier than this, then there is more potential 
energy being lost by a than is being stored by B or wasted in 
friction, and it must be stored up in some other form. The 
surplus stock shows itself in a quicker motion of everything ; 
it is being stored up as kinetic energy. 

190. We have now to consider an important question. 
When a certain amount of potential energy (measurable in 
foot-pounds) disappears, and becomes kinetic energy, how 
quickly must all the parts of the machinery move to store 
it all up*? This problem is very troublesome, because every- 
thing in Fig. 22 is in motion in a different way ; some parts 
of the mechanism are moving slowly, others quickly. It is, 
however, easy to find out how much kinetic energy a small 
body has if we know its weight and its speed. Let there bo 
a small ball hung from the point o, Fig. 162, by a silk thread, 
so that when it vibrates we can call it a simple pendulum. 



APPLIED MECHANICS. 



243 




Now, you know that when it reaches the end of its swing at a 
it is, for a very short interval of time, motionless, and has no 
kinetic energy. It falls from A to B ; and as there is almost 
no friction, we may suppose that the 
potential energy which it loses in falling 
through the vertical height from A to b 
is all stored up as kinetic energy when 
the ball reaches b. The body has a 
certain velocity in feet per second when 
it reaches b, and it is on account of its 
having this velocity that we say it has a 
store of kinetic energy. If the vertical 
height from b to a is li feet and the 
weight of the body is w lb., we have 
w h foot-pounds, which it had at A, con- 
verted into kinetic energy at b. Now 
comes the question. We know that 
this kinetic energy is proportional to w, 
but how does it depend upon the velocity? 

It evidently does not depend on the 
direction of the velocity at b, and some 
mathematicians would say that this 
shows that it must depend upon an even power of v, but 
this is metaphysics. 

Now, experiment shows that when a body falls freely at 
London without friction, its velocity when it has fallen freely 
without friction at any place through distances proportional to 
0, 1, 4, 9, 16, 25, 36, &c., is proportional to the numbers 0, 1, 
2, 3, 4, 5, 6, etc. Experiment shows that these numbers are 
quite independent of whether we measure in feet or centi- 
metres or yards, in minutes or seconds. Hence we say that 
the velocity is proportional to the square root of the height, and 
we assume that we are stating our result when we say v^ cc h, 
or v^~h= a constant, which we shall call b, ov v'^ = bh . . . (I). 

Notice that when we write our result (1) down we have 
written down something which is beyond mere arithmetic ; it 
is on the very much higher subject — physics, which in its 
mathematical form is called algebra, v, b and h are not mere 
numbers ; they are quantities, and if v and A were merely to be 
considered numbers, the dimensions being feet and seconds, the 

assumption in (1) involves v- ( "— j ) ■=hh (feet), or b = -• 



Fig. 162. 



244 APPLIED MECHANICS. 

(feet) 
(second)^' ^* ^^' ^ ^^ ^^^^ ^ ^^^^ number ; but we may take 

it to be the mere number ^ multiplied by , ^^^ „ In truth, 

A (seconds). 

we find out in various ways that h had better be written 2 g 
where ^is an acceleration of 32-2 feet ner second per second in 
London. 

We have , then, the rule for bodies falling freely from rest, 
v:= ^2gh....(l). 

Now, h is v'^l 2 g, and hence, when a body has fallen h 
feet, and its velocity is v, the potential energy, wh, lost by 

it, may be written w —- or, as it is more usually written, 

2g 
1 w 
o —V-... (2). w is the gravitational force or weight of the 

body in pounds, and g is 32-2 feet per second per second in 
London. 

191. It was by experiments on falling bodies that Galileo 
was led to formulate the law that at any place a body's w/g, 
which we call the mass or inertia of it, is constant ; and we have 
been led by astronomical observations to believe that the mass or 
inertia of a body is a constant everywhere. The gravitational 
force on a body^ w, may vary, and g, the acceleration due to 
gravitational force, may vary ; but the ratio between them 
seems everywhere to be constant. We generally denote the 
mass or inertia wjg by the letter m. A body kejDt in London, 
defined by law to have a weight of 1 lb. if weighed in vacuo, 
has a mass 1 -r- 32-2 in the absolute units which are most con- 
venient for engineering purposes. Any body whose weight in 
London is w lb. has a mass numerically expressed by w-i-32-2. 
Any body whose weight in any kind of units is w at a place 
where the gravitational acceleration in any units is g, is said to 
have a mass w/g. 

If we wish to be very exact, we notice that if we speak of 
a force as w lbs., the w is a mere number ; if we speak of a 
force as w, then w is much more than a mere number. A 
body whose weight in London is 32*2 lbs. has a mass 1, 

192. We have now seen that half the mass of a body, multi- 
plied by the square of its speed in feet per second, is its kinetic 
energy. When the bob, Fig. 162, is at b, let us say that its 
total store of eneigy is kinetic. When it is at a, the energy 



APPLIED MECHANICS. 245 

is all potential. When it is anywhere between, its total amount 
of energy is exactly the same as before, but part is potential 
and part is kinetic. During the swinging of a pendulum there 
is a constant change going on, potential energy chauging into 
kinetic or kinetic into potential, and the sum of these two 
would always remain the same only that friction is constantly 
reducing this sum by converting part of it into energy of 
another order — namely, heat. 

Exercise. — Imagine no frictional resistance to the motion 
of, a projectile. A projectile of 100 lbs., with a muzzle velocity 
of 1,000 feet per second. What is its kinetic energy when 
it is at heights of 10, 50, and 100 feef? When its whole 
velocity is 707 feet horizontally per second, what is its height ? 
This being the horizontal component of the velocity through- 
out, what was the angle of elevation of the gun % 
Arts., 1,551,000, 1,547,000 and 1,542,000ft. lbs.; 8,241 ft.; 45°. 

It is to be noticed that we assume that the potential 
energy of a body is w, multiplied by the vertical height through 
which it can fall (that is, to some fixed datum level), so that at a 
certain height we know its potential energy ; and if we know its 
total energy, the remainder is kinetic, and the kinetic being 

always - m v^", the velocity is known when the height is known. 

193. Test of our Law. — We now have our rule to calculate 
the energy stored up in a moving body, every part of which 
is moving with the same velocity, and we can test it in the 
following way. Get a pulley (Fig. 163) as light and frictionless 
as possible, because we must, at the beginning, neglect both the 
energy stored up in the jiulley itself and the loss by friction. 
Fasten the pulley at a considerable height above the floor. 
Let two equal weights, a and B, balance one another at the 
ends of a long silk cord, passing over the pulley ; and let there 
be a wooden scale, close alongside which a passes as it ascends 
and descends. Let us be able to fix to this scale, at any place, 
a plate which will suddenly stop A, and, above this, a ring 
which will just allow a to pass through. You will find such 
an arrangement as I speak of in almost every little collection 
of apparatus in the kingdom, and it is called an Attwood's 
Machine. Now let a be as high as possible at the beginning ; 
place on it a little weight w, such as will be lifted oft' when a 
passes through a ring ; and place a ring so that it will lift the 
littlf) weight oft" A when A has fallen, say, 3 feet. You know 



f46 



APPLIED MECHANICS, 




that, SO long as fche little weight lies on a, the speed of A down- 
wards and B upwards must become greater and greater. In 
fact, the potential energy lost by the 
little weight becomes converted into 
kinetic energy of the whole arrangement. 
Now, as soon as the little weight is 
stopped, A and b move with a steady 
motion ; and if the table is placed by 
trial so that one secontl after a passes the 
ring it is suddenly stopped by the table, 
the distance between the ring and table 
shows the velocity which a, b, and the 
little weight had when the little weight 
was removed. In one experiment — a 
being 1 lb. and b the same, and the little 
weight 0'25 lb. — the velocity was 
measured after a had fallen 3 feet, and 
was found to be about 45 feet per second. 
Now, the potential energy lost by the Little 
weight was 3 x 25, or '75 foot-pound. 
Tlie kinetic energy was stored up in 2-25 
pounds, moving with the velocity of 4-5 
hiet per second, and, according to the 
al:x)ve rule, its amount is 

2-25 ^ 64-4 X 4-5 x 45, 
or -71 foot-pound, or 04 foot-pound too 
little. If we consider that there was 
some friction, that the pulley retained 
some kinetic energy, and that it was 
difficult to fix the table, so that exactly one second elapsed 
from AS passing through the ring until it was stopped, we see 
that the experiment is a fairly good illustration of the rule. 
You ought, with your own hands, to make a number of such 
ex|)erinients. In exercise work we shall call a = b = w lb., 
and we shall call the little extra weight if?. 

In Ai-t. 50 we found the sort of connections which had to 
be introduced by friction at the wheel pivots and by the bend- 
ing of the cord. Their study led us to the great subject of 
friction. But in a well-made Att wood's machine, a far more 
important matter to consider is the energy (so far neglected) 
in the pulley. The experimental study of this leads us to a 
consideration of angular motion. 




P^isa 



APPLIED MECHANICS. 247 

194. Energy in a Rotating Body. — Suppose now that the 
pulley is so massive that its kinetic energy is considerable, and 
may not be neglected, is there any way of finding from its speed 
how much energy it has stored up in it i We can easily calculate 
the energy in any little portion of a wheel if we know its 
velocity and mass, but those portions near the centre are 
moving more slowly than portions near the circumference, so 
that we have to calculate the energy in each little portion 
separately, and add all the results together. There is one 
thing which all portions of a wheel have in common — they all 
go round the centre the same number of times per minute. 
Suppose now that the number of revolutions of a wheel is 
doubled, the real velocity of every point in the wheel is doubled, 
whether that point be near the axis or not, so that the kinetic 
energy of the whole wheel is quadrupled ; in fact, then, we find 
that the kinetic energy stored up in a wheel depends on the 
square of the number of revolutions which it makes per minute, 
so that the energy must he equal to a constant momher multiplied 
hy the square of the number of revolutions per minute. 

195. To find experimentally how much energy is possessed 
by a wheel when it is rotating, let the wheel be mounted on an 
axle supported on very frictionless bearings. If the centre of 
gravity of the wheel is not exactly in the axis, then it is better 
to place the wheel as in Fig. 164. Now let a cord with a 
loop from the pin b be wound round the axle and pulled by a 
weight w. Suppose the weight to be 1,000 lbs., and that we only 
allow it to fall 8 feet from rest before the cord drops off the pin, 
so that when it has fallen this distance it no longer acts on the 
wheel, which will then rotate with a constant speed. Roughly 
speaking, the wheel possesses 1,000 x 8, or 8,000 foot-pounds of 
energy stored up in it. This is not quite true, because the 
weight itself possessed a certain amount of energy of motion 
which must be subtracted. Suppose that at the instant before 
being stopped the weight was moving with a velocity of 1-5 foot 
per second, then we must subtract 

^ X 1-5 X 1-5, or about 35 foot-pounds. 

64-4 

If there were no friction, and we find that a speed of 10 

revolutions per minute has been given to the fly-wheel, we 

know that we have to find a constant number, M, which, wlien 

multiplied by the square of 10 or 100, will give 7,965 foot^ 

pounds. Evidently M = 79-65, and hence, if ever we find this 



248 



APPLIED MECHANICS. 



fly-wheel rotating, we know that it has stored up in it the 
amount of energy in foot-pounds 79-65 x square of number of 
revolutions per minute. 

196. In the above calculation we have neglected friction; 
but, as a matter of fact, in experiments the friction never is 
negligible. As for the friction of the wheel itself, on a cord 
similar to that which you have already used, hangr a small 
weight such as will merely overcome friction, so that when you 




Pig. 164. 

give the wheel a jerk for the purpose of starting the motion, 
this weight will just suffice to prevent friction reducing the 
speed. Suppose this weight to be 5 lbs., then it is quite evident 
that 5 lbs. of the original 1,000 were really employed in over- 
coming friction and not in storage. Hence our calculation gives 

995 X 8 - 35, or 7,925 foot-pounds as the total storage. 
This is at ten revolutions per minute. When it makes one 
revolution per minute the storage is 79*25 footpound?, and at 
any other speed we multiply 79-25 by the square of the number 
of revolutions per minute; 79-25 is called the M of the wheel. 

As for the friction of the pulley p and the energy wasted 
in bending the rope, the pulley must be tested like the pulley 



APPLIED MECHANICS. 249 

of Art. 60, and a correction made on account of the two 
parts of the rope being at right angles to one another. Or 
two pulleys may be tested at the same time, the cord going 
over both of them to two weights. 

197. It is obvious that we must be pretty quick in count- 
ing the number of revolutions of the wheel produced by the 
falling of the weight. Indeed, we ought to observe, if possible, 
the time taken in part of one revolution, using some special 
form of time-measurer, because the speed will now continually 
decrease on account of friction. 

But there is another way in which it is easy to find the 
speed at the instant when the weight ceases to act. Find the 
total number of revolutions made by the wheel till the 
cord drops off its pin, and let someone observe this time in 
minutes. Then, as we know that the speed increases uniformly 
during this interval of time, the mean speed is just half the 
speed at the end of the interval ', that is, divide the number of 
revolutions hy the number of minutes in which they were per- 
'formed, and twice the quotient will give the number of 7^evolutions 
per Tninute made by the wheel when the weight just ceases to act. 
You can test your result by counting the number of revolu- 
tions from the time the cord drops off until the wheel is 
stopped by its own friction and dividing by the time which 
elapses; twice this quotient ought also to be the speed you 
want to know. 

198. It is not necessary even to measure the friction 
directly, for we found that 7,965 foot-pounds were given out by 
the weight in falling; now if we count the total number of revolu- 
tions made by the wheel from the time of starting until stopped 
by its own friction, and divide 7,965 by the total number, we 
shall find the loss of energy due to friction during one revolution, 
since there is just as much energy wasted by friction in any 
one revolution as in any other [a statement to be tested by 
the student]. Ten times this must be the same amount of 
energy as 5x8, or 40 foot-pounds, for we measured the friction 
during 10 revolutions of the wheel as equivalent to 5 lbs. 
falling 8 feet. This, then, is the method we ought to employ. 

I know of no exercise in my laboratory which is so useful 
as this. A thoughtful student revives his knowledge of 
nearly all the important dynamical principles in making tho 
corrections, which enable him to get nearer and nearer to the 
correct answer for M. 



250 APPLIED MECHANICS. 

199. You see that Mis a number which ought to be known 
for every fly-wheel ; it is just as important to know the M of a 
fly-wheel as to know the weight of an ordinary body. We have 
only to multiply the M by the square of the number of revolu- 
tions per minute, and we find at once the energy in foot-pounds 
stored up in the wheel. I have shown you liow to find the M 
of a fly-wheel by experiment ; I will now give you an idea of 
its value in different cases. Imagine a grindstone whose 
diameter is 4'5 feet, whose breadth is 1*4 foot, the weight 
of its material per cubic foot being 132 lbs. ; then we can 
calculate its M by first finding 

132 X 1-4 X 4-5 X 4-5 X 4-5 X 4-5, 

and dividing this answer by 59,800. For any rotating object 
of cylindrical shape, the shape of a grindstone, this rule will 
always find M- Multiply the weight of the material jyev cubic 
foot by the breadth or width ; multiply this by the fourth power 
of the diameter, and divide by the constant number 59,800. 
Whether the material is wood or stone or metal, this will give 
M, and this multiplied by the square of the number of revolu- 
tions per minute will give the energy in foot-pounds stored up 
in the rotating body. For the above grindstone, on calculating 
out, you will find the M to be 1"26. So that when it makes 
1 revolution per minute, there is stored up in it 1-26 foot- 
pound of energy ; when it makes 2 revolutions per minute, 
there is stored up in it 1-26 x 4, or 5-04 foot-pounds. 

Foot-pounds. 

At 



3 


revolutions, 2-53 x 9, or 22-68 


20 


„ 2-53 X 400, „ 1,012 


50 


2-53 X 2,500, „ 6,325 


.00 


2-53 X 10,000, „ 25,300 



200. If we fix a small weight of 20 lbs. on a wheel, at 12 feet 
from the axis, this adds to the M of the wheel the amount 

20 X 12 X 12 -^ 5,873, or 0-49 ; 
or the weight multiplied by the square of its distance from the 
axis, divided by 5,873. 

If we add a very thin rim to a wheel, the addition to M ia 
found by multiplying the weight of the rim by the square of its 
average radius, and dividing by 5,873. 

Note. — In Table II. aU dimensions are supposed to be in feet See 
Art. 203. 



APPLIED MECHANICS. 
Table II. 



251 



Nature of Rotating 
Body. 



M 



Sphere of diara-'j 
eter d, rotating f ,5 .^^ ^„^ 
aboiit diameter h ^''^^ 001626 
as axis. . . , / 



xvd^ 4- 112,166 



•3162ci 



Spherical shell, ■ 
whose outside 
diameter is d 
and inside is c?i, , 
rotating about 
diameter as axis . 



ID [d^ - di^) ■ 
X -001626 



^ 112,166 



3i«^^^ 



Cylinder, diam- " 
eter d, length I, 
rotating about 
its axis . . . . 



ivld* X -00305 



tvld'i -r- 59,814 



•3535d 



Hollow cylinder, 
outside diameter 
d, inside diam- 
eter di, length I 



lOl [d^-di^] 
X -00305 



wl{d^-di^) 
-T- 59,814 



3535v/(Z2+c?]2 



Thin rim, mean \ 
radius r of > 
weight W . . j 



wr2 -s- 32-2 



wr2 ^ 5,873 



Thin rod, of 
length I, rotat- 
ing about axis 
through its 
middle point, 
at right angles 
to its length. 
Weight of rod W 



Thin rectangular^ 
plate, rotating 
about axis 
through its cen- 
tre parallel to 
side 6, the side 
d being at right 
angles to axis. 
Weight of plate 
w 



WZ2-00258 



wd? -00258 



WZ2 ^ 70,474 



wcZ2 ^ 70,474 



-2882(i 



252 APPLIED MECHANICS. 

It will be found that if a fly-wheel has light arms and 
a heavy rim, as we often see on such wheels, a fairly good 
approximation to its M is found by multiplying the weight of 
the wheel by the square of the mean diameter of the rim^ and 
dividing by 23,000. 

Example. — The rim of a fly-wheel weighs 15 tons ; its mean 
diameter is 20 feet. Calculate approximately what energy is 
stored up in it when it makes 60 revolutions per minute. Here 
you will find the M of the fly-wheel to be about 584, and hence 
the stored energy is 584 X 60 X 60, or 2,102,400 foot-pounds. 

The mathematicians do not use our M. They use a 
quantity called I, the moment of inertia of a rotating body, 
which is numerically equal to twice the energy stored in the 
body when it has an angular velocity of one radian per second. 
See Art. 203. I give the values of I as well as of M in the 
above table. 

In the case of any rotating body, if we imagine a fly-wheel 
made of the same weight and same M or I, with an exceed- 
ingly thin rim, the average radius of this rim is called the 
radius of gyration of the body. In fact, the mass of a body 
multiplied upon the square of its radius of gyration is its 
moment of inertia. The radius of gyration of a fly-wheel rim 
is usually taken to be the average radius of the rim. 

201. Steadiness of Machines.— A fly-wheel is put upon a 
riveting- or shearing-machine, or other machine, because the 
supply of energy to the machine is not given regularly, or else 
because the demand for energy from the machine is irregular. 
The fly-wheel enables the machine to maintain a more constant 
speed. In calculating the proper size of a fly-wheel for any 
machine we must know two things : first, what is the greatest 
alteration of speed allowable in the case ; and secondly, the 
greatest fluctuation of the demand and supply of energy. Thus, 
suppose we wish never to have the speed of the fly-wheel 
more than 51 nor less than 49 revolutions per minute, and 
that during some interval of time the fly-wheel has to give 
out 20,000 foot-pounds more than it receives during that time; 
then, although the fly-wheel will afterwards have this deficiency 
made up to it by some steady supply, it is obvious that 
its speed must diminish. We wish its speed to diminish 
only from 51 revolutions to 49 revolutions per minute in 
this interval of time. Now, when the fly-wheel runs at 51 
revolutions, it has stored up an amount of energy equal to 



APPLIED MECHANICS. 253 

its M X 51 X 51; and when it runs at 49 revolutions, its 
store is M X 49 X 49, and the difference between these two 
ought to be 20,000. Hence, subtracting 49 x 49, or 2,401, 
from 51 X 51, or 2,601, we get 200; and dividing 20,000 by 
200, we find 100 as the required value for M. Subtract, then, 
the square of the least speed from the square of the greatest, and 
divide the greatest excess of demand or supply by this remainder ; 
the quotient is the M of tlie fly-wheel. Having found M, the 
question is, how can you tell from it the size and weight of the 
wheel'? Find the M of any wheel of the same shape and 
material as that which you want to use. It is obvious that the 
diameters of the wheels are as the fifth roots of their M's.* We 
want a wheel whose M is 100. Suppose I find a wheel of the 
shape I wish to use whose outer diameter is 8 feet, and I 
calculate its M, and find it to be 11; then 

The fifth root of 11 : fifth root of 100 :: 8 : answer. 

Log. 11 = 1-0414; divided by 5 it is 0-2083, which is the 
logarithm of 1-615. 

Log. 100 = 2-0; divided by 5 it is 0-4, which is the 
logarithm of 2*512. Hence 

1-615 : 2-512 :: 8 : answer. 

This is an easy exercise in simple proportion. I find my 
answer to be 12-44 feet, or 12 feet 5 J inches, the diameter of 
the required fly-wheel, which is to be similar in form to the 
smaller specimen used by me for calculation. 

202. The total kinetic energy stored np in any machine 
is found by calculating the energy in every wheel and in every 

* If we liave any two similar wheels, or other rotating bodies of the same 
material ; if we consider any similar small portions of them ; it it evident that 
their weights are proportional to their cubic contents, or to the cubes of any 
similar linear measurements. Hence, if one is, say, twice the diameter of the 
other, as every dimension of the one is twice that of the other, the weight of 
one must be 2 X 2 X 2, or eight times that of the other. Now, the M of any 
rotating body depends not merely on the weight of each portion of the body, 
but on the square of its distance from the axis, so that the M of one must be 
8x2x2, or thirty-two times the M of the other. Similarly, if the linear 
dimensions were as 3 to 1, the values of M would be as 243 to 1 for a pair of 
similar wheels. 

Example. — "We want a wheel which will have a store of 1,000 foot-pounds 
when rotating at twenty revolutions per minute, and it is to be of the same 
shape as that of an already existing wheel, which is 4 feet in diameter, and 
which contains a store of 1,350 foot-pounds when running at 30 revolu- 
tions. Evidently the M of this second wheel is 1,350 -r- 900, or 1*5, and the 
M of the first wheel is to be 2*5. Using logarithms, we find that the fifth 
loot of 15 is to the fifth root of 2-5 as 4 feet is to 4-4 feet, the answer. 



254 APPLIED MECHANICS. 

moving part, and adding all together. But suppose that in the 
machine there is some shaft of more importance than any other, 
it is usual to give the speed of this shaft only, because if its 
speed be doubled, the speed of every other is doubled. Thus, in 
a steam-engine we state the number of revolutions per minute 
of the crank shaft, and this tells us the speed of every part of 
the engine. Let, then, the number of revolutions of some 
such principal axle of a machine be found. If this number 
of revolutions is doubled, the kinetic energy stored up in the 
machine is quadrupled ; and, in fact, the kinetic energy stored 
up is equal to a certain number vjhich can he found for the 
machine, and which we shall call its M, multiplied hy the 
square of the number of revolutions of this particular axle jyer 
minute. The M of any machine may be experimentally deter- 
mined in exactly the same way as we have shown above. 

If we know the M of any machine, then the M of any other 
machine made to the same drawings, and of the same materials, 
but with all its dimensions twice as great, is thirty-two times 
as great, because the M^s of the two machines are proportional 
to the fifth powders of their corresponding dimensions. 

203. The energy stored up in a rotating "body is equal to fia^, 
wtere I is moment of inertia about the axis ; that is, the sum of all 
such terms as mass of a little portion multiplied hy the square of it« 
distance from the axis, a is angular velocity in radians per second. 

Hence, as o = -^ , if « is number of revolutions per minute, and 

IT is 3 '14 16, the energy is i ; so that our M is ~ In 

^^ 1,800 1,800 

Table II. we give values both of M and of I. In both cases w 

is in Ihs. per cuhic foot, and all dimensions are supposed to he in 

feet. 

We ask for the advice of students in an important matter. Is 
it good to use the idea of our M, the energy stored in a body when 
it makes one revolution per minute ? To the weaker vessel, the 
beginner, or the man who dislikes algebra, we know it to he 
useful. Indeed, to any practical engineer, however mathematical 
he may be, it is convenient to have such a quantity. But is its 
convenience overpowered by the inconvenience of having a new 
quantity to think of ? We do not know, and we should like to 
receive ad^dce. 

As for the idea of moment of inertia, it comes in in this way. 
If we have a small mass m moving round an axis at the distance 
r feet with angular velocity a, its linear velocity is ra feet per 
second ; its kinetic energy is ^mr^a^ foot-pounds. Now, if we have 
many small masses and make this calculation, we see that every 
httle term has ^a^ in it, common to them all. Hence we multiply 



APPLIED MECHANICS. 255 

every little m by its r^ and add up, calling the sum I, or, as we 
say mathematically, Smr^ = i, and evidently fia^ is the kinetic 
energy. In any hook on the calculus, exercises will he found on 
the calculation of the various values of I given in Tahle II. 
Every student who knows how to do so ought to calculate all 
these, and also the value of I for various other bodies in such 
shapes as may be defined mathematically. 

EXERCISES. 

1. A fly-w^heel of cast iron, whose rim is 10 feet mean diameter and 
section 8" x 10", has a volume 8xl0xl07rXl2 cubic inches, and as 
a cubic inch weighs 0'26 lb. the weight is 7,840 lbs. Its moment of 
inertia is, approximately, 7,840 x 5'^ -h 32-2, or 6,087. Find its M also. 

The M of a fly-wheel is its kinetic energy when going at one revo- 
lution per minute. This is an angular velocity a = 2ir -j- 60, or 0"1047 
radians per second ; so that M is ^ x 6,087 X (•1047)2 = 33-48. 

2. What energy will this fly-wheel store in changing from 98 to 102 
revolutions per minute? Ans., m (102^ — 98"), or 26,782*8 foot-pounds. 

3. A gas engine has 6 indicated horse-power at 150 revolutions per 
minute ; what is the indicated work of one cycle (or two revolutions) ? 

Ans., 6 X 33,000 -f- 75, or 2,640 foot-pounds. 

4. If one-half of this is stored in changing speed from 149 to 151 
revolutions per nJnute, what is the M of the fly-wheel ? What is its 
moment of inertia ? If it is made to the same drawings as the fly-wheel 
of Exercise 1, what is the average radius of its rim ? What is its weight ? 

Ans,, 2-2 ; 400; 2-9 ft. 1,534 lbs. 

5. If the gas-er.gine of Exercise 3 goes at 200 revolutions per minute, 
and if there is a complete cycle in every revolution and half of its energy 
is stored, what is the M of the fly-wheel if fluctuation of speed is to be 
the same as in (4) ? A97S., 0-46. 

6. In Attwood's machine each w = 1-5 lb., w = 0-3 lb. If w is 
applied for a height of 2 feet, what ought to be the velocity ? 

Afis., Total mass is 3-3 ~ 32-2, or 0-1025, and its kinetic energy is 
^ X -1025 X ^2. This is equal to 0-3 x 2, or 0-6 foot-pound. Hence 
v2 = 1-2 _-^ 1,025, or «; = 3-42 feet per second. 

7. If in Exercise 6 the wheel weighs 0'17 lb., with a radius of gyration 
which happens to be equal to the radius of the circle of the centre of the 
cord as it goes round, we have simply to imagine the moving mass to 
have 0-17 lb., or rather 017 -^ 32-2, or -0053, added to it. What is the 
corrected value of v at the time when w is lifted off ? 

Ans., The whole mass is now -1078, and ^ x -1078 v^ = 0*6 ; so 
that V = 3-34 feet per second. 

8. If the radius of the cord circle on the pulley is 0-24 foot and the 
radius of gyration of the pulley is 0-21 foot, imagine the pulley of 
0-17 lb. and radius of gyration 0-21 replaced by another of x lbs. and 
radius of g;y'ration 0-24 ; so that x x (0-24)2 = 0-17 x (0-21)2, ^^ 
X = 0-13. We therefore imagine a mass 0-13 -;- 32*2, or -0040, added on 
to the original weight. In this case find «;. 

Ans., V = 3-36 feet per second. 
It is this sort of calculation, when experimentiag with an 
Attwood's machine, which gives a man a practical knowledge of 
mechanics. 

9. A homogeneous cylindei of mass m and radius r roUs down an 



256 APPLIED MECHANICS. 

inclined plane. If the linear speed of its centre Ls v, show that its 
angulai velocity o about the centre is vjr. What is its kinetic energy ? 

Ans., ^?nv'- -\- |ia-. 
Table I. shows that its radius of gyration, k, is -doSod ; and 
AS 1 = mk-, we have the kinetic energy ^n {v'^ + k-a^), or 

^m fv^ + F !lV or ^mv- (l + -TV Xotice that in a cylinder 

k'-/r- = i ; so that the total kinetic energy is Smv-/^. If the 
cylinder is cast iron, r = 0-17 foot, its length 0-3 foot ; then its 
weight is 12-257 lbs., its mass is 0-3808, and its kinetic energy 
is 0-2855v2^ 

If this cylinder has rolled without friction along a switchback 
path through a vertical height of 1-5 foot, show that i? = 8-02. 

If the path was an inclined plane, such that for a vertical faU 
of 1-5 feet the distance traversed is 20 feet, and if the average 
velocity was half the final velocity, find the time taken to reach 
this point. Ajis., 4-99 seconds. 

On a plane at the angle a with the horizontal, show that 
the time taken for a sloping distance I feet, being ^ -^ ^, is 
Vol a sin a. 

10. A riveting-machine needs 3 horse-power; a fly-wheel upon it 
fluctuates in speed between 80 and 120 revolutions per minute; an 
operation occurs every two seconds, and this reqiiires fths of all the 
energy supply for two seconds. Find the M and I of the fly-wheel. 

The energy supply for one minute is 3 x 33,000, and l-ths of the 
supply for two seconds is 3 x 33,000 x f x -^ x 2, or 2,875 foot- 
pounds. This is equal to m (120- — 80^) ; so that M is 036, I is 
•36 X 1,800 -^ 7r2, or 65'6. 

11. A machine (consisting of a g-rindstone and the shafting of a shop) 
has an M = 12-13. Suppose the loss of energy by friction in one 
revolution to be always 1,550 foot-pounds, whatever the speed. If the 
machine (the grindstone spindle) is making 140 revolutions per minute 
and ceases to receive energy, in how many revolutions wlLL it come 
to rest? 

Let the answer be x ; then x x 1,550 = 12-13 x 140-, so that x = 154 
revolutions. As the average speed will be 70 revolutions per minute, 
the stoppage will take 2-2 minutes. 

12. Find the M and I of a fly-wheel which stores 25,000 foot-pounds 
of energy when changing in speed from 50 to 70 revolutions per minute. 
It is to be similar, and of the same material, to an existing fly-wheel 
whose M is 0-047. What is the ratio of their sizes ? 

Ans., 10-4; 189-7; 2-93: 1. 

13. If the earth described a circular path of 92 x 10^ miles radius in 
365J days, if it were a sphere of 8,000 miles diameter, of uniform density 
0-6 times that of water, what is its mass in engineer's units ? What is 
its kinetic energy because of the motion of its centre? What is its 
kinetic energy because it rotates on its axis once in 23 hours 56 minutes 
t seconds ? 

14. An engine is running at 200 revolutions per minute. Suppose 
that after the steam is shut off and tbe load removed the engine made 
250 revolutions before coming to rest. Assume a constant moment of 
resistance, and calculate its amount if the moving pai-ts are capable of 



APPLIED MECHANICS. 257 

storing as much energy as a fly-wheel weighing 1 ton and having a radius 
of gyration of 2| feet. Ans., 73-6 lbs. 

15. The rim of a fly-wheel is 9 inches broad and 4^ inches deep, the 
external diameter heing 9 feet. Find the moment of inertia of the rim. If 
an engine be supposed to drive this wheel (neglecting other resistances), 
how many revolutions will have been made before the speed acquired from 
rest by the wheel is 120 revolutions per minute, the diameter of the 
cylinder being 18 inches, the stroke 2 feet 6 inches, and the mean effective 
pressure 15 lbs. per square inch? What time would be required ? 

Ans., 2134 ; 17-6 revs. ; t = 8-8. 

16. The fly-wheel of an engine of 4 horse-power, running at 75 revolu- 
tions per minute, is equivalent to a heavy rim, 2 feet 9 inches mean diameter, 
weighing 500 lbs. Estimate the ratio of the kinetic energy in the fly-wheel 
to the energy developed in a revolution, and determine the maximum and 
minimum speeds of rotation when the fluctuation of energy is one-fourth 
the energy of a revolution. Ans., 1 : 1*94 ; 75-2, 74-8. 

17. A fly-wheel is made to rotate by means of a weight of 5 lbs. 
attached to the end of a cord passing round the shaft, the diameter of 
which is, together with thickness of cord, 1 J inches. Find the moment 
of inertia of the wheel if the weight falls a distance of 8 feet from rest 
in 1 minute, friction being neglected. Ans., 4*35. 

204. Exercise. — In the Attwood's machine, if the pulley of moment of 
inertia I is suj)ported on foin- friction wheels each of moment of inertia I^ 
about its own axis, and if the spindle of the j)uUey has a diameter one- 
twentieth of the diameter of the rim of each friction wheel, show that the 
pulley has a ^drtual moment of inertia I -f "01 1^. 

In a good Attwood's machine calculate carefully the virtual moment 
of inertia of the pulley. Find experimentally for any values of a and b 
the lessening of the preponderating force which represents the fiiction 
and stiffness of the cord. If you can work with a discarded old machine 
with massive wheels and much fiiction and hemp-cords, you will find it 
far more instructive than an expensive, nearly frictionless contriA^ance. 
In measuring time, exercise a little ingenuity of your own in getting 
accm-acy ; but if yoiu- teacher has made everything very easy for accurate 
measurement, you will learn nothing. If you have read too much about 
other people's methods, you will learn very little. You will learn most 
when you try to get accuracy of measiu-ement with very rough apjjaratus. 
Teachers who do not think will introduce roughness and inaccuracy in 
the wrong places ; they will give you a bad timekeeper, perhaps, or an 
inaccurate scale for measuring distances. 

After having worked with the Attwood's machine, in which the velocity 
ratio is 1, you ^\dll find it very interesting (if you have time enough) to 
experiment with the apparatus shown in Fig. 22, in which the velocity 
ratio may ha-s'e any value. In the experiments described in Art. 51 you 
aimed at keeping speed constant. Now let there be an excess weight b, 
and measure the kinetic energy as with the Attwood machine. We need 
not here describe more fully the obviously interesting and instructive 
experiments which may be made. 

205. In a Bull engine {see Fig. IG 6) steam-pressure in the 
space B, 2^ lb. per unit area (in excess of the pressure from above 
the piston) acts on a piston of area A with a total upward force/) A. 



258 



APPLIED MECHANICS. 



This causes a total weight w to be lifted (consisting of steam- 
piston, rod, pump-plunger, etc.). If p varies when the piston 
rises through a height h, the total work e done by the steam must 
be calculated for small changes of level, and the results added 
up. Thus in Table IIa. we give the pressure of the steam as 
the piston rises, tiie pressure being measured by an indicator. 
The area of the piston is 900 square inches, so that 900 p is 
the lifting force. It is therefore evident that the fourth column 
shows at each place (approximately) tlie work already done 
upon the piston when w (w = 20 tons, or 44,800 lbs.) has 
risen h feet. If it is rising with the velocity v feet per 

w 

second, it possesses w/i foot-pounds of potential and |— -y" foot- 
pounds of kinetic energy. But we know the total e ; and if 
we subtract wA, we know the kinetic energy, and we can 
therefore calculate v. 

TABLE IIa. 





p Pounds 




E, the total 




1 w „ 




h Feet. 


per Square 


A j3 Pounds. 


Wort done in 


wft. 


^-v\ 


V. 




Inch. 




Foot-pounds. 




y 







100 


90,000 














0-5 


100 


90.000 










1-0 


100 


90,000 


90,000 


44,800 


45,200 


8-06 


1-5 


100 


90.000 










2-0 


100 


90,000 


180,000 


89,600 


90,400 


11-4 


2-5 


80 


72,000 










3-0 


66-7 


60,030 


252,000 


134,400 


117,600 


13-0 


3-5 


57-1 


51,390 










4-0 


50-0 


45,000 


303,390 


179,200 


124,190 


13-36 


4-5 


44-4 


39,960 










o-O 


40-0 


36,000 


343,350 


224,000 


119,350 


13-1 


5-5 


36-4 


32,760 










6-0 


33-3 


29,970 


376,110 


268,800 


107,310 


12-41 


6-5 


30-8 


27,720 










7-0 


28-6 


25,740 


403,830 


313,600 


90,230 


11-38 


7-5 


26-7 


24,030 










8-0 


25 


22.500 


427,860 


358,400 


69,460 


9-98 


8-5 


23-5 


21,150 










9-0 


22-2 


19,980 


449,010 


403,200 


45,810 


810 


9-5 


21 


18,900 










10-0 


20 


18,000 


467,910 


448,000 


19,910 


5-33 


10-25 


19-5 


17,550 










10-5 


19 


17,100 










10-75 


18-6 


16,740 


481,600 


481,600 









APPLIED MECHANICS. 



259 



206. The earnest student will work out the numbers in 
the above table, and draw curves showing h and|? and h and v. 
In Fig. 165 OGHIJ shows jo the pressure, o f being the distance 
travelled by the piston and weight. The ordinates of o p N repre- 
sent the velocity at every instant. Kotice that if we divide the 




c c 


B 


B 


w 





Fig. 165. 



Fig, 166. 



force pK and the weight w by a, wc get/) and w/A to compare. 
The straight line k m represents w/a. The area of o K m n is 
equal to the area o G H u N ; that is, w/a is the average value 
of the pressure until A = ON. . The student must see clearly 
that the force increasing the velocity is represented to scale by 
the ordinate of kghlijmk, and this force becomes negative 
after L, so that the point l tells us where v reaches a maxi- 
mum and begins to diminish. The positive area k G h l k is 
equal to the negative area l u m L. It is worth while spending 
a good deal of time over such curves and such an investigation 
as this. 

The student ought now to assume that besides the force 
w there is a constant force of friction (say 2,000 lbs.) to be 
overcome. He will therefore subtract 2,000 h foot-pounds 

w 

from E, as well as w A, to find the kinetic energy, 1 - v^, and 

he will calculate v, again obtaining a new curve. In this 
engine, when the steam is allowed to escape, the weight w 
falls, and performs the actual pumping operation. 

207. Shoiild the advanced student care to take up a problem, that 
more nearly approaches the actual case (namely, assume that the 
force of friction r is not constant, but is some function of the 
velocity), he must find out for himself some graphical method of 
working. Our own plan is this : — We calculate v at the end of an 
interval from a rough notion of the velocity, and therefore of f, the 



260 APPLIED MECHANICS. 

force of friction during the interval. We can now approximate 
more closely to the actual friction and calculate more exactly. 
We ad^dse the student to proceed in this way in the above case, 
taking f as following the law 

r = 500 + 100 V. 

When a torque M turns a body through the angle radians, it 
does work of the amount m 9. Thus, if a turning moment of m 
pound- inches is acting on a shaft which revolves at n revolutions 
per minute, or 2Tr}i radians per minute, it does work m (pound x 
inch) 2Trn j)er minute, or tt?? m -r 6 foot-pounds per minute. 

If the twisting moment on a shaft m is proportional to 6 
radians, the angle through which a certain length of shaft is 
twisted, then |m 6 is the strain energy. 

If bending moment m acting on a certain length of a beam 
between two cross-sections causes an angular change 0, and 6 is 
proportional to m, then |m 9 is the strain energy produced by the 
action of m. These two propositions enable us to calculate the 
resilience of many springs ; that is, the energy which it is possible 
to store up in them. 

The force p, acting through a little distance 5x, does work 
p . 5a?. If F varies, a curve must be drawn showing its value for 
each valae of as, and the work done in any distance is shown as an 
area or an integral. Thus, if f = ax, it is easy to show that the 
work done from a; = to any other value is lax^ ; or from x = x-^^ 
to a? = 072 t^® work done is ^a [x.^ — x-^). Or we may put it that 
the work done is x multiplied by the average value of f, the average 
value being the half sum of the two extreme values. 

Thus if the gradually applied load tv on a beam produces the 
deflection y where y is proportional to w, the energy stored in the 
loaded beam is \wi 

But if there is no loss of the energy due to loading, any method 
of loading which during the deflection oj is calculated to do the 
same work will cause the same deflection. One such method 
is to suddenly apply half the final load or \w. Here, as in 
structures generally, we have assumed that deformation is pro- 
portional to load. If the law of jd elding is force =f{y) where 
there is any curious law, and if the integral of f{y) with regard to 
y is r(?/) ; or if in case the deformation is through an angle Q the 
moment m =f{Q), then the energy stored in any configuration is 
y[ij) or r(0). If a force or moment which, gradually applied, 
would only produce a yielding ^ or 9 is suddenly apj^lied of its 
full amount, the yielding is y^ or Q^ where y^f{y) = F(yi) or 

01/(0) =f(01). 

Thus if the righting moment m of a ship is a known function of 
the heeling angle Q (say m = a sin. aQ), if the moment m due to 
the wind produces the steady heeling angle 9, and the energy 

stored in the heel is — I a- cos. a9.dd:= - (1— cos. a9), if this 



'-j: 



moment were suddenly apjilied, and the vessel heeled over under a 
gust of wind from o to 6^, the work done is 0jM or 0jA sin. ad ; and 



APPLIED MECHANICS. 



261 



as this is all stored as - (l - cos. aO-^), we can calculate 6^, knowing 

e from Oi A sin. aO = - (1 - cos. aO-^^). The work is easily done 

gTaphically. If the 
curve o p R V repre- 
sents the righting 
moment (ap) corre- 
sponding to the in- 
clination e (or TP), 
then a steady mo- 
ment p Q produces the 
heel s if the rectan- 
gular area o t p u s o 
is equal to the area 
of OPRSO. If the 
law is M = A sin. 40, 
the vessel cannot 

right herself if is more than 45 degTees. But if the student 
studies the matter, he will see that if a steady moment gets the 
vessel over to anything approaching 45 degrees, she must heel 
farther than 45 degrees. We are neglecting friction. 

There is the same moment at $2 as at d when sin. aO^ = sin. a9 ; 

and if 6^ sin, a0 = - (1 - cos. aO^), the steady wind which would 

keep the ship at 9 will, as a steady gust acting from = 0, heel 
the vessel to 0^ and heyond. Thus, taking the above example 
a = 4, 




sin. 40„ = sin. 40, or 402 = ■"■ 



40, 



or 0, = ^- 



02 sin. 40 = i (1 - cos. 40o), 
^ COS. 402 = COS. (tt - 40) = - cos. 40, 

(t - 0) sin. 40 = i (1 + cos. 40), cos. 40 = 2 cos. 2 20 



(^ - 40) 



sm 
COS. 20 



40 



sin. 20 
^ _ 40 — cot. 20 = 0. 
Calling this (p (0)=O, I find that =llf 
degrees. Hence, if a wind is such that 
it would maintain a steady heel of llf 
degTees ; if it caught the vessel sud- 
denly and acted steadily, and we ne- 
glect friction, the heel would he as 
much as 33J degrees ; and if the 
wind still continued to act with the 
same moment, the righting moment 
heing now less than the moment due 
to the wind, the vessel must go on her 
heam-ends. She will recover from 
a gust of wind less strong than this. 



= cot. 20, 



0deg. 


e 


,^(0) 


18° 
16° 

12° 
11° 


•3142 
•2793 

•2095 
•1920 


0-508S 
0-4241 

•0576 
- -1015 



262 



A.PPLIED MECH.^XICS. 



208. So long as a constant force acts it produces a uniform 
acceleration in the direction in which it acts. We may experi- 
ment with Attwood's machine, or simply use, as the body acted 
on, a small carriage running very freely on a very smooth level 
table; and the force acting, the pull in a string passing over a 
pulley on the edge of the table, and having weights in a 
scale-pan at its end, Fig. 168. Here, however, friction will 




Fig. 168. 

be greater than in Attwood's machine, 
We can illustrate the following rule. If 
a force of 2 lbs. acts on a body whose 
weight is 50 lbs. at London (the 50 lbs. includes the 
weight of everything which is set in motion, so that if we 
use a little w^eight of 2 lbs. for the purpose of exerting the 
force, we must remember that this little weight of 2 lbs. is 
included in the 50 lbs. ; we may take 48 lbs. as the body 
acted upon, and the pull in the string, which is less than 
2 lbs., as the force), then the acceleration or increase of the 
velocity every second is equal to the force divided by the whole 
mass moved. In this case the mass is 50 -f- 32-2, or 1-553, so 
that we have 2-i-l'553, or 1*223 feet per second per second as 
the acceleration. Thus, if the body started from rest, the 
velocity would be 1*223 x 5, or 6*115 feet per second at the end 
of 5 seconds. And now comes the question, how far will the 
body move from rest in five seconds ? Evidently its average 
velocity during this time is half its terminal velocity, or 3-058, 
so that 5 X 3-058, or 15-388 feet is the distance. It is evident 
that to get the sj^ace passed over we have multiplied half the 
acceleration by the square of the time. 

I do not suggest that the apparatus of Fig. 168 is the most 
suitable apparatus for use in the laboratory ; there is too 
much friction, and it seems ditFtcult to measure the velocity. 
Attwood's machine is used for this illustration, and is described 



APPLIED MECHANICS. 263 

in Art. 193. In both pieces of apparatus corrections must be 

made for the motion of wheels. 

"When a body falls freely, its own weight is acting on its 

own mass. For instance, say the weight is 2 lbs., then the 

mass is 2 -f- 32*2, and weight divided by mass is acceleration, 

which we find to be in every case 32-2 feet per second per 

second at London. Anywhere else than in London its weight 

is w lbs., and its mass is, as before, 2 -i- 32-2 ; and consequently 

2 
its acceleration is w -^ ^^ . But we call this acceleration en 
32-2 -^ 

and hence w = 2 x j^ We see that — anywhere is the 
32-2. g 

same as anywhere else. Keeping to London, where g =: S2'2 

feet per second per second, the velocity at the end of any 

number of seconds is 32-2 multiplied by this number ; and the 

space fallen through in any number of seconds is half 32*2, or 

16-1 multiplied by the square of the number of seconds. You 

can check these rules by the rules given you for potential and 

kinetic energy, and you will find them quite consistent with 

one another. 

209. Momentum. — If a body's weight is 2 lbs., its mass is 
2 ~ 32-2, or •062L Now, if the body is moving with a velocity 
of 20 feet per second, its momentum is -0621 x 20, or 1'242. 
If this momentum is created or destroyed by a force acting for 
only one second, the force must be 1*242 lbs. ; if it is created or 
destroyed by a force acting for 5 seconds, the force must be 
1-242 -^ 5, or '2485 lb. The mass of a body multiplied by its 
velocity represents its momentum. Momentum is sometimes 
defined as the quantity of motion possessed by a body. It 
represents the constant force which, acting for one second, 
would stop the body. Suppose a certain amount of momentum 
is produced by a force of 1 lb. acting on a body for one 
second, the same amount of momentum would be produced by 
a force of 2 lbs. acting for half a second, or by 1,000 lbs. acting 
for the thousandth of a second, or by '001 lb. acting for 1,000 
seconds. 

Example. — A bullet of 2 ounces, or -125 lb., at 500 feet per 
second, directed towards the centre of mass of a body of 200 lbs. 
at rest, in which it lodges and which is free to move ; what is 
the velocity after the impact *? The momentum before impact is 

— -— X 500. The momentum aftei impact is ---, — '^- x the 



264 APPLIED MECHANICS. 

. , , . ^ . . -125 X 500 

required veiocitv. Hence, the answer is — ^„^ ■, ^_ , or 

0"312 feet per second. 

Fxercise.—Chii^i^ing hammer, IJ lbs. ; a velocity of 30 feet 
per second is destroyed in the one tive hundredth of a second. 
What is the average force of the blow ? 

Ans., 700 lbs. (nearly). 

Example. — One hundred and twenty cubic feet of water leave 
the rim of the wheel of a centrifugal pump everyminute; itscom- 
pouent velocity in tlie direction of motion of the rim is 25 feet 
per second. ^Miat is the retarding force on the vanes at the 
rim of the wheel ] — Ans., Two cubic feet of water per 

second have the mass - "" '^ ] this, multiplied by 25, gives 

the momentum lost by the wheel per second, which is 
force, and amounts to 97 lbs. If the velocity of the rim 
is 30 feet per second, the work done per second is 30 x 97 
foot-pounds; the work done per minute is 30 x 97 x 60; 
dividing by 33,000, we find 5-56 horse-power. Assuming that 
there is no force at the inside of the wheel, the water entering 
radially and without shock, this is the power given to the 
water. If we neglect friction inside the wheel and also out- 
side it retarding its motion, this is the total power given 
to the wheel itself. 

What is the work done per pound of water ? There are 
2 X 62-4 lbs. of water per second, and the work done per second 
is 30 X 97 foot-pounds, so that the work done per pound of water 

30 X 97 

^ ^ ?^-T~Ti 01' 23 "3 foot-pounds, or enersv sufficient to lift the 

2 X 62-4' ^ ' 

water to a height of 23 -3 feet. 

210. If the velocity of a body has been produced or destroyed 
by various forces, each acting for a certain time, multiply each 
force by the time during which it acted (each of these products 
is called an impulse), and the sum must be equal to the 
whole momentum generated or destroyed. When we know 
the time during which a certain force has acted on a body 
giving to it motion, we generally determine the motion by 
calculating the momentum of the body. When we know the 
distunce through which a force has acted on a body giviug to it 
motion, we generally first find the kinetic energy of the body. 

211. Knowing the force f acting at any instant on the mass 



APPLIED MECHANICS. 



265 



M, the acceleration a is f -^ m. Thus, suppose the following 
values of f to be given ; the varying force acting on the mass 
of a body whose weight is 644 lbs. in London. In engineers' 
units its mass is 64*4 -^ 32-2, or 2. Suppose that at time the 
body has a velocity 2;=30 feet per second. 



Time, 
in seconds. 


in pounds. 


in feet per 

second 
per second. 


in feet per 
second. 


in feet. 





20 


10 


30 





•1 


20 


10 


31 


305 


•2 


19 


9-5 


31-970 


6-199 


•3 


18 


9 


32-900 


9-443 


•4 


16 


8 


33-750 


12775 


•5 


14 


7 


34-500 


16-188 


•6 


11 


5-5 


36-125 


19-669 


•7 


8 


4 


35-600 


23-205 


•8 


5 


2-5 


35-925 


26-781 


•9 


2 


1-0 


36-100 


30-382 


1-0 


-1 


-0-5 


36-125 


33-993 


1-1 


-3 


-1-5 


36-025 


37-601 


1-2 


-4 


-2-0 


35-040 


41-163 



To obtain the numbers in column 4, take an example. 
Suppose we know that when t='4: second from the beginning 
v=33'750 feet per second. Now, in the next O'l second, the 
average acceleration is approximately J (8+7), or 7*5; and in 
the time 0*1 the actual increase of velocity is 7*5 x 0*1, or "75 ; 
and this is what we add to 33-750 to get 34*500 the velocity at 
the end of the little interval. 

We warn beginners that there is no easier way than this, 
of getting several very important, essential ideas, and every 
number of such a table ought to be calculated. Now, notice 
that s, the space passed over, is made up by multiplying an 
interval of time 0*1 by the average velocity during tha.t 
interval. Thus, at ^ = '4, s = 12*775 feet is the distance 
passed through since the time ^ == 0. During the interval from 
^ = *4 to ^ = -5 the average velocity is ^ (33*75 + 34*50), or 
34*125 feet per second, and the space passed through in this 
0*1 second is 34*125 x 0*1, or 3*4125 ; and this, added to 12*775, 
gives 5=16*1875, or, rejecting the last figure, s= 16*188 at the 



2Q6 



APPLIED MECHANICS, 




time^=0'5 seconds from the beginning. Note that in approximate 
calculations of this kind we cannot be certain of our last figures 
in each number. It will assist the student now to illustrate 
this work bv curves. Plot a and t so that the ordinate of b c D is 

a and the abscissa is t. 

L 
I 



Anyone who thinks a 
little must see that 
the area of B c E o re- 
presents (to scale) the 
total increase of velo- 
city at the time repre- 
sented by E. Add 
this on (taking care 
about the scale of 
measurement) to 30, the velocity at o, and we have the true 
velocity at the time o E. Let the velocities be found and 
plotted as G H IJ. In the same way the area of the v curve 
must give the space s curve ; that is, the area of o g h e o 
represents to some scale the space s passed over since the time 
o, and we can now show on a curve s at every instant. This is 
given as o K L. 

To repeat : e c represents to scale the acceleration at the 
time e ; E H represents to scale the velocity at the time o E ; 
it is the area of o b c e o added to the velocity at the time 0, 
care being taken as to the scale of the diagrams ; E K represents 
s ; it is measured as the area of G H e o. 

A student who works such an exercise as this carefully is 
getting all sorts of valuable notions, not merely of mechanics 
but of practical mathematics. Unfortunately, twenty academic 
exercises can be worked out without thought or trouble to 
teacher or student, and by the rules of the game this is 
sufficient for the passing of examinations. For the present, 
therefore, my advice will be followed by a few earnest students 
only — the men who want to know, the men who are not merely 
in search of examination tips, the men who find academic 
exercises difficult because they think about what they do. 
If such exercises as the above ever become obligatory on all 
examination candidates, of course their academic friends will 
discover ways of doing such exercises without the trouble 
of thinking about them. 

212. When the resultant force f acting on a body of mass m is 
constant, the acceleration a=F -r M is constant. There is no such 



APPLIED MECHANICS. 267 

case in Kature, but it is commonly studied. Wlien a body falls in 
a vacuum in an ordinary laboratory, and there are no magnetic 
or electric or frictional effects, we may for all practical purposes 
assume that the force, the weight of the body w, is constant. 
The mass is w -^ ^, and the constant acceleration is ^, or 32-2 
feet per second per second in London. If the student treats this 
case in a table like that of p. 265, or by means of curves like 
Fig. 169, he will see ih-dt v = g t, s = ^ g t"^, and hence that 
-yS — 2 ^ s. Or if Vq is the velocity downwards at the time 
t = and s is the vertical height through which the body 
falls from ^ = o to any other time t, then v = Vq -\-gt, s^VQt + 
yf-, a.iid2gs = v^--VQ^. . ; . 

If a body w lbs. falls without friction down an inclined 
plane, making an angle a with the horizontal, the constant 

w 
force is w sin. a, the constant acceleration is w sin. a -f- - o r 

9 

g sin. a. 

In any case when the acceleration a is constant, v = v^^-rat 
s = %t -^ ^at% 2as = v^ — v^^^. 

213. I have described the units of measurement employed 
practically, not merely in calculation but in thought, by 
English-speaking people. In some parts of our work we find 
it necessary to calculate in a system based upon other units — 
the centimetre a,s the unit of length, the inertia or mass of the 
amount of stuff called one gramme as the unit of mass or 
inertia, and the second as the unit of time. This is called the 
C.G.S. system. Its advantages lie in its being used by scientiiic 
men of all countries. One of its disadvantages lies in this, 
that all answers to problems must be multiplied by coefficients 
to bring them into practical language (see p. 655). 

EXERCISES. 

1 . A bullet takes 2^ seconds to fall to the bottom of a well. What 
are the depth and the velocity at the bottom ? Assume no resistance of 
the atmosphere. 

Ans., Depth s = Ig {2\)- ; and taking g = 32-2, s = lOOG feet. The 
velocity is 2|^, or 79 o feet per second. 

2. The bullet of (1) leaves the hand with a velocity of 20 feet per 
second downwards. What are the two answers ? 

s = 20 X 21 + Ig (2|)2 = 1506 feet ; also «; = 2|(7 + 20 = 99-5 
feet per second. 

3. The hullet of (1) leaves the hand with a velocity of 20 feet per 
second upwards. What are the two answers ? 

s = _ 20 X 2| 4- i^ (2i)2 = 50 6 feet; v = 2^^ - 20 = 59-5 feet 
per second. 



268 APPLIED MECHAK1C8. 

4. How liigh. did the bullet reacli in (3) ? 

Ans.,v^^ == 2ffh, or 20^ -i- 2^ = A = 6 21 feet. 
In the above exercises time and space are measured from leaving 
the hand. 

5. A bullet leaves o, a point on the ground, with an upward velocity 
of 300 feet per second. Find i/, the vertical height of it, at the times 
^ = 0, ^ = -1, ^ = 2, etc., seconds. 

6. A bullet leaves o with a horizontal velocity of 400 feet per second, 
and no force acts upon it to alter its horizontal velocity. Find x, its 
horizontal distance from o, at the times 0, -1, -2, *3, etc., seconds. 

7. If a bullet has both the velocities of (5) and (6) when leaving o, 
plot its positions on a sheet of squared paper at the times 0, -1, '2, etc., 
and show that the path is parabolic. 

8. With different horizontal and vertical components, but the same 
total velocity (500 feet per second), let the bullet of (7) leave o and again 
plot the path. Do this in several cases. If you knew a Kttle mathe- 
matics, you could prove that an angle of elevation of 45 degrees will 
give the greatest range on a horizontal plane. 

9. A force acts on a body of 8 ozs. for 6-912o minutes, and produces a 
velocity of 10 feet per second. Find the force. Express it in dynes 
(see p. 655). 

Ans., 0000372 lb., or 165*6 dynes. 

10. How far will a lateral force of 1 oz. m^ove 100 lbs. on a smooth 
horizontal plane in 5 minutes ? Ans., 9i)0 feet. 

11. In Attwood's machine, where the weights are 17 ozs. and 16 ozs., 
find the acceleration and the tension of the cord. 

Ans., 0*976 foot per second per second ; 1 lb. 

12. How long must a force of 14 lbs. act on a body of 1,000 tons to 
give it a velocity of 1 foot per second? Ans., 5,000 seconds. 

13. A rifle 5 feet above a lake discharges a bullet horizontally, which 
strikes the water 400 feet away. What was the velocity of the bullet ? 

Ans., 720 feet per second. 

14. A man weighing 168 lbs. is standing on the floor of a lift. What 
force does he exert on it (1) when the lift is stationary? (2) when it is 
falling freely — that is, with an acceleration of 32'2 feet per second per 
second ? (3) when it is descending with an acceleration of 12 feet per 
second per second ? (4) if it is ascending with the latter acceleration r In 
each case indicate by a figure what are the forces which act on the man, 
and give the resultant force in the direction of motion. 

Ans., (3) 105 lbs., (4) 231 lbs. 

15. A jet of water having a sectional area of 12 square inches, and a 
velocity of 16 feet per second, impinges normally on a fixed plane surface. 
What is the mass of the water which comes on the plane per second ? What 
is the momentum of this quantity before impact ? What is the force 
on the plane ? If the jet impinges normally on a plane surface which has 
a velocity of 6 feet per second in the direction of the jet, what is the 
velocity of the water relatively to the surface ? And what is the force 
exerted on the surface ? Find the amount of work per second necessary 
to maintain the jet, and the work done by it per second ; and find the 
efficiency. Ans., 2-58 ; 41-28 ; 41-28 lbs. ; 10 feet per second : 16-1 lbs. ; 
332 ft. lbs. ; 96-7 ft. lbs. ; -29. 

16. The head of a steam-hammer weighs 50 cwts. ; steam is admitted 



APPLIED MECHANICS. 



269 



on the under side for lifting only, and there is a drop of 5 feet. What 
will he the velocity and momentum of the head the instant hefore the 
blow is given if the fall is without resistance ? If the time during which 
the compression of the iron takes place he -^ second, find the average 
force of the hlow. Ans., l7'9o feet per second ; 3,121 ; 111-5 tons. 

17. A hody has its velocity diminished hy one-third. By how much 
are its kinetic energy and momentum diminished? If this diminution 
was hrought ahout by a certain constant force acting on the hody through 
a distance of 5 feet, through what further distance would this force have 
to act in order to hring the body to rest? If, on the other hand, the 
diminution of velocity had taken place in five seconds, what additional 
time woxild be required to bring the '^odj to rest, the same constant force 
still acting? Ans., 4 feet; 10 seconds. 

213. Force may be defined as the space-rate at which work is done 

or any form of energy converted into another, or it may be defined 

as the time-rate of transference of momentum. 

We would advise students to make two sets of curves from 

Table IIa., p. 265. The fii'st is given in Fig. 169. The second set 









Area = 








Distance. 
s 


Pressure. 
P 


Acceleration. 


2 


V 


5^ 


t 





100 


32-5 











•5 


100 


32-5 






•2481 




1 


100 


32-5 


32-5 


8- 6 




•2481 


1-5 


100 


32-5 






•1027 




2 


100 


32-5 


65-0 


11-4 




•3508 


2-5 


80 


19-5 






-082 




3 


66-7 


10-9 


84-5 


13-0 




•4328 


3-0 


57-1 


4-7 






•0758 




4 


50 


•1 


89-2 


13-36 




•5086 


4-5 


44-4 


- 3-5 






-0756 




5 


40 


- 6-3 


85-7 


13-1 




•5842 


5-y 


36-4 


- 8-7 






-0784 




6 


33-3 


-10-7 


77-0 


12-41 




•6626 


6-5 


30-8 


-12-3 






•0841 




7 


28-6 


-13-7 


64-7 


11-38 




•7467 


7-5 


26-7 


-14-9 






•0936 




8 


25 


-16-0 


49-8 


9-98 




•8403 


8-5 


23-5 


-17-0 






•1106 




9 


22-2 


-17-8 


32-8 


8-1 




•9509 


9-5 


21 


-18-6 






•1490 




10 


20 


-19-3) 
-19-6 -9-8 

-19-9( 
-20 j -4-9 


14-2 


5-33 




1-0999 


10-25 


19-5 






-1217 




10 5 


19 


4-4 


2-97 




1-2216 


10-75 


18-6 


-0-5 





•1700 


1-3916 




1-3916 





270 APPLIED MECHANICS. 

ought to have s for the horizontal co-ordinates or abscissae. We have 
seen that it was a most instructive problem, when given the a, t 
curve, to find the v, t and s, t curves. Xow, suppose we have the 
a, s ciu've and we wish to find the v, s and the t, s curves. As a 
matter of fact, we have already worked out a v, s curve when given 
an a, s curve in Art. 211. But let us look at it from our new point 
of view. In the Bull engine of Art. 205 the force causing upward 
motion of w is Ap — w ; the mass is w/y, and so the acceleration 

is a= Cap — w) -H ^, or a = ( p — ^] -i- —. This accelera- 
ff \ aJ Ag 

tion is given in Column 3 of the table on p. 269 ; Column 1 shows s, 

and fi'om s and a we can draw our 5, a curve, which is really shown as 

GHLiJMKof Fig. 165, a being the ordinate measured upwards from 

-vT dv dv ds dv J , , , 

K M. Now, a=— = -— . _ = v, and hence a . ds = v . dv, or 
dt ds dt as 



!■ 



ds = Iv^, or twice the area of the a , s diagram up to any 



place is the square of the velocity. 

We have, in our usual way, worked the integral of a . ds 
numerically in the table, and we give the values of v. 
ds 

Note that as v = -^, or 5^ = Ss/v, we get the intervals of time 

by dividing the intervals of space by the average velocity during 
the interval. Thus the interval of time from s = 4 to s == 5 feet 
is 1 foot H- i (13-36 + 13-1), or 0-0756 second; and if we had 
already determined that t = '5086 for s = i, we know that 
t = -5086 + -0756 for 5 = 5. In this way the numbers of 
column 6 were obtained. 

If the V, s diagTam is given, and we are asked to find the a, s 
diagram, we notice that 

dv dv ds dv 

''~di~"d^'dt~^'ds ^ '' 

Therefore a is rejDresented by the sub-normal to the v diagram. We 
advise no student to use the measurement of the normal of a ciu-ve 
for any useful piu']30se. It is practically too inaccurate. But 
from a table of the values of v and s, taken from a curve which 
will correct errors of observation, values of Sv/Ss, and therefore of v 
5v/ds or a, may be calculated with no great error. The values of t 
may be obtained as in the last example, and the whole motion is 
known. 

Of course if of the variables t, s, v, «- any one is known as an 
algsbi^ic function of the other, it is an exercise in the calculus to 
find an} or all the others in terms of any one ; as also the work 
done, or the kinetic energy, and other things. 

Example. — A body has reached the earth at London from space, 
no other force than the earth's gravitation having acted upon it ; 
what is its velocity ? If r^ feet is the distance of London fi'om 
the earth's centre, and at any other place reached by the body if 

the distance is r, then g, the acceleration at r, is gQ-%^- Let 2gQrJ^ 
be called h^, then *"^ 



APPLIED 5IECHANICS. 271 

<7 = ^ — , and - — = —«' = — *_,... (1). 

_ dr dh' y^ dr ,, , /^/•\2 , }fi 

2 j7 ■ :^ = 2 • x» so tliat ( j- ) or v^ — + _ . . _ (2). 

dt dt^ t^ dV \dt) r ' 

dr 

Let — = when r =. co, and we need to add no constant. Hence 
dt 

-r = -, and tlierefore i-^dr = — b . dt. We take the — siffa 

dt r* ° 

as we imagine the hody falling. Integrating again, we have 
f r 2 = — bt + c. Let ^ = when ?• = r^, so that all onr times 
before reaching the earth will be negative. 

|,.J = , ...^=^^(,,^1 _,.!).... (3). 

It is (2) that we asked for. 

Note that a body of w lbs., when it reaches r from space, has 

w b^ 
a kinetic energy ^ , or w r. We are therefore prompted to 

study the problem from the energy rather than the momentimi 
point of -sdew. Imagine a body of 1 lb. leaving the earth (say at 
London). 

At the distance r feet from the earth's centre the weight of the 

body is -|- if Vq is the distance of the earth's surface from the 

centre. The potential energy here being v, and being y + 5v 

at r + dr, 5v = weight x 5r = -| . 5r. Hence v = ^ + 

const. 

If Vq is the potential energy of 1 lb. (we are really in all this 
neglecting the fact that the earth moves relatively to the object) at 

the earth's surface, Vq = ^ -}- const., so that our constant is 

Vo -\- Tq. Hence v = - -^ + Vq + ^o- 

The potential energy when r = oo is v„ + r^. Hence we have 
the easily remembered fact : — A body of w lbs. lifted infinitely high 
from the earth's sm-face would receive a store of w r^ foot-pounds of 
energy if r^ is the radius of the earth in feet ; and if we imagine 
all this converted into kinetic energy, we see that the velocity of 
the body comiug from rest in space would be \^ 2g^r^ when ^o = 
32*2. If g is the gravitational acceleration at the place r, the 



272 APPLIED MECHANICS. 

tody's speed on reacliing this place would be v' 2gr ; and if we 
remember that g = yo'"o7''^j calculation is easy, v = i/ Ig^r^^lr. 
If we write 'o =■ — ji where t is time, and if \/ 2(j^r^ is called 

h, then — r\ . fir = b . cU, or — ^r^ = bt -^ const. Thus, if we 
count time from the time of reaching the earlh's surface, so 
that it is always negative, let ^ == when r Tq. The constant 

_2_ 

Also, as vrl == b, ^2 = -^, and hence ^ = ^ (^'o^ 3), or 

125. Force F is rate of change of momentum m. If force f acts 
for time 5^, it increases the momentum of a body by the amount 5 m ; 

so we can say either that f = -r-, or that I y . dt = whole gain 

of momentum if t is the time during which the gain occurs. 
Of course if f is constant, we have f t = gain in momentum ; 
or the whole momentum gained is the force multiplied by the 
time. But if f varies, we can only say that the whole gain of 
momentum, divided by the time, is the average force duiing the 
time. Here we have a time average. 

TOO work done 

Space average value of force = ^^,,1^ distan ce' ^^"^^ ^^'^^^' 

„ „ momentum gained 

value of force = ., , ,. . 

whole time 

Continually in dynamics we are considering the two great ideas 
of energy and momentum. On any system if a force acts from 
outside bodies it gives energy, and it gives momentum. If no 
force acts from the outside, the momentum and moment of 
momentum remain unaltered ; and the total energy would remain 
unaltered were it not that other forms of energy become changed 
to heat, and a system loses heat by radiation. If the earbh-moon 
system were alone in s^Dace, we have to consider that its moment of 
momentum remains constant, whereas its total store of mechanical 
energy is diminishing. Professor Purser pointed out that this idea 
gives us the past and future history of the earth-moon system, and 
Professor Darwin has worked it out for us. 

One of the most instructive of laboratory experiments is that 
in which two bodies, a and b, are suspended so that they may 
collide, their motions before and after collision being measurable. 
The whole momentum after impact is the same as before impact ; 
and it is very interesting to notice that when a strikes b at rest 
the extent of swing of the two in combination after impact is not 
affected by the initial rebounding and chattering and the many 
little circumstances which cause the energy after impact to be quite 
different from what it was before. Men who do not make 
experiments of this kind have no clear notions of dynamical 



APPLIED MECHANICS. 



273 



phenomena, unless they are very exceiDtional (that is, men of 
genius). 

When the momeutuin m, as of a jjile-driver, is destroyed in the 
time T, m/t is the time average force of the blow ; we have a 
perfectly definite idea of what is meant. But when we are told 
that the whole energy of the falling pile-driver, divided by the 
distance through which the pile is forced into the groimd, gives us 
the resistance of the gi'oimd to the pile, we get a misleading 
academic statement. The blow is a complicated phenomenon ; and 
even in the much simpler case of the history of the collision of two 
billiard-balls we are only now beginning to see how and where the 
total energy is converted into heat (see Art. 404). 

Indicator diagrams of engines are space diagrams of force. 
Their average heights enable the work done to be calculated. 













It 






t 


F 


^ a 


- a5^ = 
- Iv 


V 


V — 


s 


P. §5 














8-8 








•1 


49 


-141 


-028 


8-756 


1-760 




86 


•2 


97-9 


-281 


-112 


8-772 




1-76 




•4 


195-8 


•562 




8-687 


3-475 




680 


•6 


293-8 


•844 


•337 






5-24 




•8 


391-7 


1-125 




8-350 


3-334 




1,310 


1-0 


489-6 


1-406 


•562 






8-53 




1-2 


587-0 


1-687 




7-787 


3-115 




1,830 


1-4 


678-4 


1-968 


•787 






11-69 




1-6 


783-4 


2-250 




7-000 


2-800 




2,194 


1-8 


881-3 


2-540 


1-016 






14-49 




2-0 


979-2 


2-812 




5^984 


2-393 




2,350 


2-2 


1077-1 


3-193 


1-277 






16-93 




2-4 


1175-0 


3-374 




4^707 


1-883 




2,220 


2-5 


1224-0 


3-519 












2-6 


1175-0 


3-374 


1-350 






18-85 




2-8 


1077-1 


3-193 




3-357 


1-353 




1,460 


3-0 


979-2 


2-812 


M25 






20-22 




3-2 


881-3 


2-540 




2-233 


•893 




790 


3-4 


783-4 


2-250 


•900 






21-16 




3-6 


678-4 


1-968 




1-333 


•533 




362 


3-8 


587-5 


1-687 


•675 






21-74 




4-0 


489-6 


1-406 




-658 


-263 




129 


4-2 


391-7 


1-125 


-450 






22-04 




4-4 


293-8 


■844 




-208 


•083 




24-5 


4-6 


195-8 


-562 


•225 


- -017 




22-16 




4-8 


97-9 


-281 


•028 


-055 


-•007 




— '7 


5-0 











--045 




22-19 






8-844 


21-892 


13,635 



274 



APPLIED MECHANICS. 



Wlien a mass is vibratiiig at tlie end of a spiral spring, the 
space diagram of the force exerted "by the spring upon the hody is 
a straight line. The space a^'erage of the force (neglecting the 
weight I'f the hody) hetween the end of a swing and the mid 
position is half what the force was at the end of the swing; 
whereas the time average of the force in this interval is the 

2 
fraction — of the force at the end. 

TT 

216. Example. — A hody of 5 tons moving at 6 miles per hour; 
what are its momentum and kinetic energy? Find the time 
average of the force which will stop 'it in 5 seconds. — Ans., The 
mass is 347"8 ; momentum, 1-97 x lO'^; the kiaetic energy is 
13466-8 ; time average of force = 612 lbs. 

If the force increases for 2^ seconds, and then diminishes again, 
in both cases uniformly with time, draw curves showing the 
velocity with time, and also with distance ; also of force with 
distance. "What is the space average of the stopping force ? 

Draw a curve showing acceleration a and t {ais negative) ; the 
integral of this shows v and t, the ordinate at ^ == o being 6 miles 
per hour, or 8-8 feet per second. The integral of the v, t curve 
shows s and t where s is distance from where the force began to 
act. Now a X mass represents force. Hence the values of a 
represent force to some scale. Xow plot a new curve showing 
force and s. The whole area of it is the kinetic energy. The 
average height of it is the space average of the force. — Ans., 621 lbs. 

We have performed the integrations numeiically, and have 
shown the results in the table on p. 273, and we have shown the 
various curves in Fig. 169a. 




169a. 



o A A A B shows a and i, t parallel to o d. 
o c E E B shows V and t, t parallel to o d. 
o 1 1 J B shows s and t, t parallel to o d. 
o H H H G shows F and s, s parallel to o d. 
o c v v G shows V and s, s parallel to o d. 



275 
CHAPTER XII. 

MATERIALS USED IN CONSTRUCTION". 

217^ Mere reading will give to no student a knowledge of 
the properties of materials. I insist on the necessity for work 
in a pattern-shop and a fitting-shop and forge ; and setting work 
in machine tools. Workshops at a college or school are not 
intended for the teaching of trades, which can only be done in 
real shops beside real workmen. A student leai-ns facts about 
materials which are necessary for his study of mechanics ; it is 
a secondary matter that he also acquires some skill which 
enables him to learn his trade quickly in a real shop afterwards. 
Town boys buy their toys and never come in contact with 
nature ; country boys are always making things and learning 
much besides the properties of materials. 

218. Stone. — The rocks which have once been melted, and 
have cooled slowly, are usually hard, compact, strong, and dur- 
able. They are most easily worked when regard is paid to the 
fact that they naturally divide up into certain regular shapes. 
They are all more or less crystalline in texture. Stratified 
rocks are those which have been deposited at the bottom of a 
sea or river ; they are often easily divided in a direction 
parallel to the layers of which they are built up, but sometimes 
there are lines of easy cleavage in other directions. These 
rocks vary very much in appearance, according to the method 
of their formation, and to the heat and pressure to which they 
have been subjected, sometimes being very crystalline, strong, 
and durable, like inarhle. Slaty rocks may be hard and durable, 
or soft and perishable ; they are not much used in construction, 
except as roof covering. Sandstones are hardened sand of 
very different degrees of compactness, porosity, strength, and 
durability. There are limestones whose particles seem to form 
one continuous mass, and which, when they have been subjected 
to great heat and pressure, become marbles; there are also 
limestones which are composed of distinct grains cemented 
together, and which may vary very much in compactness, 
strength, and durability. Besides these there are conglomerates^ 
in which fragments of older rocks are imbedded. A little 
knowledge of geology is necessary in order to understand the 
properties of rocks. Stones are preserved by coating them 



276 APPLIED MECHANICS. 

with some material such as coal-tar, various kinds of oil and 
paint, or soluble glass, which fills their pores and prevents the 
entrance of moisture. An artificial stone, which can be made 
in blocks of anj recjuired size and shape, is obtained bv turning 
out of moulds and afterwards saturating with a solution of 
chloride of calcium, a mixture of clean sharp sand and silicate 
of soda. The chloride of calcium and silicate of soda produce 
silicate of lime, which cements the sand together, and thus 
gradually consolidates the whole mass. 

219. Bricks. — Bricks are made of tempered (that is, freed 
from pebbles, saturated with water, and well ground and mixed) 
clay, moulded, dried gently, then raised to and kept at a white 
heat in a kiln for some days, and cooled gradually. Sand in 
the clay prevents too much contraction and helps vitrification. 
Bricks should have plane parallel surfaces and sharp right- 
angled edges, should give a clear ringing sound when struck, 
should be compact, uniform, and somewhat glassy when broken, 
free from cracks, and able to absorb not more than one-fifteenth 
of their weight of water. They ought to require at least half 
a ton per square inch to crush them. The published tests are 
sometimes much more than a ton per square inch. Probably 
half the published strengths are the true strengths of bricks or 
of brickwork. The standard brick is 8| x Ij X 2| inches. 
The average weight of brickwork is 116 lbs. per cubic foot. 
A bricklayer lays from 100 to 200 bricks per hour. 

220. Limestone, when burnt in kilns, gives ofi" carbonic 
acid. If pure it forms quick-lime, which combines readily with 
water, becoming larger in volume. Mixed with clean sand this 
forms mortar, which, in the course of time, hardens by losing its 
water and combining with carbonic acid from the air. If the 
burnt limestone were not pure, but contained certain kinds of 
clayey materials, u-on, &c., it would not combine with much 
water, but when ground up fine^ water enables its particles to 
combine chemically with one another, forming compotmd sili- 
cates with gi^eater or less rapidity, depending on its composition. 
Such cement first sets, acquiriug a large degree of firmness, and 
then more slowly and without much expansion becomes as hard 
as many Limestones. These natural hydraulic limestones are 
not much used now. Nearly pure limestone or chalk is mixed 
with about one-third of its volume of blue clay to produce — 
when ground and mixed in plenty of water, then drained and 
dried, then burnt to incipient vitrification and ground up again 



APPLIED MECHANICS. 277 

very finely indeed — an artificial cement, which is equal, if not 
superior, to the natural cement. This is the Portland cement 
now in use. Fineness in the particles is exceedingly important. 
Sand in mortar saves expense, and prevents the cracl^iing ot 
the mortar in drying; coarse sand seems better than fine. 
Two measures of sand to 1 of slaked lime or 3 to 7 of sand to 
1 of cement are the average allowances, but every person v/ho 
uses mortar ought to test a particular lime or cement to see 
how much sand it will bear to have mixed with it. Concrete 
is a mixture of 6 of gravel or broken stones and 1 of cement. 

From the time of setting the tensile strength of cement 
increases at first rapidly and gradually more slowly. The 
French standard is 280 lbs. per square inch at the end of seven 
days, 500 lbs. in 28 days, 640 lbs. in 84 days. 

The initial strengths of neat cement, 1 of cement to 2 of 
sand, 1 of cement to 5 of sand being about 1 : 5- : f ; the 
gains of strength in a year are about as 5 : 4 : 3 . Using too 
much water weakens cement ; water about i to ;^ of the weight 
of the cement is found to give the best results in testing. The 
true crus?iing stress of cement is probably about six times the 
tensile stress. 

221. Timber. — A tree is made up of a great number of little 
tubes and cells arranged roughly in concentric circles, one circle 
for each year of growth, because the sap which circulates out- 
side is checked every winter. The process of seasoning consists 
in uniformly drying the timber. As each little portion dries, 
it contracts and becomes more rigid, and it contracts much 
more readily in the direction of the circular arrangement of the 
tubes than it does towards the centre of the tree, and least 
easily in the direction along the tree. It is obvious, then, that 
if the tree is dried whole, there will be a tendency to splitting 
radially. If the tree is cut up before drying we can tell the 
way in which the planks will warp if we remember the above 
facts. 

Firwoods are easily wrought, and possess straightness in 
fibre and great resistance to direct pull and transverse load, 
and are largely used because of their cheapness. They differ 
greatly in strength, but their weak point is their inability to 
resist shearing. The best of these is the red pine or Memel 
timber from Russia, which can be had in large scantlings, and 
thus used without trussing. The white fir or Norway spruce 
is suitable for planking and light framing, and is imported 



278 APPLIED MECHANICS. 

from Christiania in "deals," "battens," and "planks." Larch 
is a very strong timber, hard to work, and has a tendency to 
warp in drying, and is therefore not suitable for framing, but 
is largely used for railway-sleepers and fences, because of its 
durability when exposed to the weather. Cedar lasts loug in 
roofs, but is deficient in strength. 

The Eng-lish oak is the strongest and most durable of all 
woods grown in temperate climates, but is very slow-growing 
and expensive. Its great durability when exposed to the 
weather seems to be due to the presence of gallic acid, which, 
however, in any wood corrodes iron fastenings ; trenails or 
wooden spikes should be used instead. Teak, which is grown 
in the East, is the finest of all woods for the engineer. It is 
very uniform and compact in texture, and contains an oily 
matter which contributes greatly to its durability. It is used 
specially in ship-building and railway carriages. Mahogany is 
unsuitable for exposure to the weather, but it has a fine appear- 
ance and is not likely to warp much in drying. It is chiefly 
used for furniture and ornamental purposes, and to some 
extent in pattern-making. Ash is noted for its toughness and 
flexibiUty, and a capability of resisting sudden stresses of all 
kinds, which make it specially adapted for handles of tools and 
shafts of carriages. It is very durable when kept dry. It is 
not obtainable in large scantlings, and is sometimes very 
difficult to work. Elm is valuable for its darabiliby when 
constantly wet, which makes it useful for piles or foundations 
under water. It is noted for its toughness, though inferior to 
oak in this respect, as also in its strength and stitfness. It is 
very liable to warp. Beech is smooth and close in its grain. 
It is nearly as strong as oak, but is durable only when kept 
either very dry or constantly wet. It is very tough, but not 
so stiflp as oak. (See also Table YII.) 

The best time for felling timber is when the tree has 
readied its maturity, and in autumn when the sap is not 
circulating. We want to have as little sap in the timber as 
possible, and in order to harden the sapwood, some foresters 
ara of opinion that the bark should be taken oflP in the spring 
before felling. After timber is felled, it is well to square it by 
taking off the outer slabs. 

Timber is for the most part dried by putting it into hot- 
air chambers, from one to ten weeks according to the thickness. 
Even when kept quite dry, ventilation is necessary to prevent 



APPLIED MECHANICS. 279 

dry rot. The circumstances least favourable to the durability 
of timber are alternate wetting and drying, as in the case of 
timber between high and low water mark, whereas good 
seasoning and ventilation are most favourable conditions. The 
most effective means adopted for preserving timber is by 
saturating it with a black oily liquid called creosote. The 
timber is placed in an air-tight vessel, and the air and moisture 
extracted from its pores as far as possible. The warm creosote 
is then forced into these pores at a pressure of 170 lbs. per 
square inch. In this way timber may be made to absorb from 
a tenth to a twelfth of its weight of creosote. 

I shall give few numbers here for the strength of timber. 
Tests of small specimens are not to be relied upon. The time 
of felling, the duration of drying, the part of the tree from 
which the specimen is cut, and many other circumstances affect 
the strength. A beam will sometimes break with a long con- 
tinued load only about half of what will fracture it in the 
ordinary way. The ultimate shearing stress along the grain of 
ash is about 600, oak 850, pine and spruce 300 lbs. per square 
inch. For bending, the average value of /in (1) of Art. 341 
seems to be for spruce 5,000, yellow pine 7,300, oak 6,000, and 
white pine 5,000 lbs. per square inch ; their Young's moduli 
being respectively 14 x 10^, 17 x 10^, 13 x 10^, and 11 x 10^ lbs. 
per square inch. The crushing strength of timber may roughly 
be taken as from 4,000 to 3,000 lbs. per square inch, and its 
tensile strength as about 10,000 lbs. per square inch. 

222. Glass. — Glass is a combined silicate of potassium or 
sodium, or both, with silicates of calcium, aluminium, iron, lead, 
and other chemical substances. Certain mixtures of flint and 
chemicals are melted in crucibles, formed when hot into the 
required shapes, and cooled as slowly as possible. This may be 
called the devitrification of glass by slow cooling, giving rise 
to crystallisation. The more slowly and more uniformly the 
cooling is effected, the more likely is it that the glass will be 
without internal strains. When glass is suddenly cooled, as 
when a melted drop falls into water, the outside is suddenly 
contracted, becomes hard and brittle, and there are such 
internal strains that if the tapering part be broken or scratched 
at the point the whole drop crumbles into a state of dust. A 
blow or scratch on the thick part produces no such effect. 
Heating and gradual cooling destroy this property. Many 
peculiarities in the behaviour of metals when heated and cooled 



280 APPLIED MECHANICS. 

seem to be caricatured in glass, possibly because tliey are due 
to the fact that all the portions of matter which are about to 
form one crystal must be at the same temperature, and when 
the substance is a bad conductor of heat there is great variation 
in temperature. Pure metals are good conductors, but the 
admixture of small quantities of carbon and of gases hurts 
their conductivity. Toughened glass is the name wrongly 
given to the hardened glass produced by plunging glass, in a 
nearly melting state, into a rather hot oily bath. This glass 
is somewhat in the condition of the glass in a Rupert's drop. 
It is so hard that it is difficult to cut it with a diamond, but if 
the diamond cuts too deep the whole mass breaks up into little 
pieces. Objects made of it may be thrown violently on the 
floor without breaking. 

223. Cast Iron. — Certain chemical changes occur when tho 
ores of iron, generally oxides, are smelted with coke ; the iron 
ceases to be in combination with the oxygen, and appears in 
metallic form, associated, however, with carbon derived from 
the fuel. There are usually other impurities besides, derived 
from the same source or from the ores. When the carbon is 
all combined with the iron the cast iron is "white," and is 
very hard and brittle. When only a little is combined, and 
most of its particles crystallise separately, the cast iron is grey 
in colour ; it is weaker and more fusible. Using the common 
names for the different varieties, No. 1 is darkest in colour, and 
from No. 4 to No. 1 there is a gradual darkening in colour. In 
the cupola of the foundry a little purification is effected, and it is 
found that the composition of a casting is from 97 to 95 per 
cent, of iron, the remainder being nearly pure carbon, often 
very largely in the combined form owing to the elimination of 
one of the impurities — namely, silicon. Nos. 2, 3, 4 are com- 
monly used in the foundry, mixtures being made of them in 
various proportions according to circumstances. A greater 
proportion of No. 3 or No. 4 gives greater strength, whereas a 
greater proportion of No, 1 gives greater fluidity and a better 
power of expanding at the moment when the metal solidifies, 
so that the sharp corners of the mould are better filled. 
Higher numbers than 4, as 8, 7, 6, and 5 — the wliite varieties 
— are seldom used in the foundry, but they may be converted 
into grey varieties by cooling from a very high temperature 
at a slow rate, but much more easily and immediately by 
the addition of certain brands of cast iron containing special 



APPLIED MECHANICS. 



281 



impurities such as " siliconeisen " or "glazed pig." A special 
degree of fluidity and resistance to action of acids is conferred 
upon cast iron by the presence of a little phosphorus, but this 
impurity renders iron fragile at low temperatures, just as the 
presence of much silicon will render it weak and liable to 
fracture from shock. To SOften a hard casting, it is heated in 
a mixture of bone-ash and coal-dust or sand, and allowed to 
cool there slowly. 

The density of cast iron varies from 6-8 in dark grey 
foundry iron to 7 '6 in white iron. Of late years cast iron has 
greatly improved in strength; this is due probably to our 
better knowledge. Contracts are sometimes undertaken to 
deliver iron of nearly 50 per cent, greater strength than the 
average number of our table, p. 411. The crushing fracture 
usually makes an angle of 56 degrees with the axis of a test 
column. The strengths of little round columns of lengths equal 
to from one to three diameters are much the same, but shorter 
columns are very much stronger, and longer columns are very 
much weaker. The reason for this is given in Art. 256. The 
specified test for cast iron is often this — that a bar 3 feet 
between supports, section 2 inches deep and 1 inch broad, should 
carry a middle load of 25 to 35 cwt., and will deflect before 
fracture 0*2 to 0-5 inch. The average ultimate shearing stress 
is 15 tons per square inch as tested by torsion. Remelting im- 
proves the strength, but not beyond a certain number of times, 
a tenacity of 6 tons per square inch in the pig becoming 9 tons 
after the second melting, and 12 tons after the fifth. This 
seems connected with the change of the iron from grey to 
white by increase of the combined carbon and decrease of 
silicon. 

Mr. Turner has arrived at the following percentages as 



giving the following 


qualities in the highest degrees : — 




Combined 
Carbon. 


Graphitic 
Carbon, 


Silicon. 


Softness 

Hardness 

General strength 

Stiffness 

Tensile strength 
Crashing strength 


0-15 
0-50 

Over 1-0 


3-1 
2-8 

Under 2-6 


2-5 
Under 0-8 

1-42 

1-0 

1-8 
About 0-8 



282 APPLIED MECHANICS. 

224. Patterns of objects are usually made in yellow pine 
(sometimes of metal if many castings are wanted), about one- 
eighth of an inch per foot in every direction larger than the object 
is to be, because the iron object contracts to thisextent in cooling. 
Gun-metal contracts about one-eleventh of an inch per foot, 
and brass about twice as much. A thoughtful pattern-maker 
can often greatly diminish the labour of moulding. Prints are 
excrescences made on the patterns to show in the mould where 
certain cores are to be placed. These cores are made of loam 
or core-sand in core-boxes, which the pattern-maker supplies ; 
they represent the spaces in the object where the melted metal 
is not to flow. They are coated with a wash of charcoal dust 
and clay. Common casting is green-sand ; there is the more 
elaborate dry-sand for such objects as pipes, and there is the 
most expensive loam moulding, in which the mould is built up 
without a pattern. You must see for yourself in a foundry 
what are the usual methods of preparing a mould : How the 
pattern is made so as to draw out easily ; how the surface of 
the sand is blackened ; how the moulder arranges his vents to 
let gases escape from the more compact parts of the sand ; 
how he places his gates to let the metal run into the mould 
with just enough rapidity and yet without hurt to the mould. 
You must also see for yourself, taking sketches in your note- 
book and making a drawing of the cupola, how the pig iron is 
melted and poured into the moulds ; how the moulder stands 
moving an iron rod up and down in one of the gates, producing 
just so much circulation and eddying motion in the melted iron 
as is likely to remove bubbles of gas w^hich may otherwise be 
unable to escape from the sides and corners of the mould, as 
well as to prevent the formation of cavities by shrinkage or 
"piping"; how in some castings he exposes to the air certain 
parts which Tvould otherwise cool too slowly for the rest of the 
object; how next morning he screens his sand and wets it. 
You ought to observe the appearance of the castings before 
and after they are cleaned up next morning. 

225. The Cooling of Castings. — The most important matter 
in connection with moulding is that there shall be the same 
amount of contraction at the same time in every portion of the 
mass of metal as it cools ; otherwise, when finished, there may be 
internal strains, which very much weaken the object, and often 
produce fracture. In designing the shape of an object which is 
to be cast, care is taken that when a thin portion joins a thick 



APPLIED MECHANICS. 283 

one it shall do so by getting gradually thicker, and not by an 
abrupt change of size. The thin piece exposes more surface, 
and cooling is effected through the surface. The thin rim of a 
pulley cools sooner than the arms, and becomes rigid sooner ; 
when the arms cool they contract so much as sometimes to 
produce fracture near the junction. In a thick cylindric object 
the outer portion becomes rigid first; now when the inner 
portion contracts it tends to make the outer portion contract 
too much, and the outer portion prevents the inner from 
contracting so much as it ought to, so that the outer portion 
retains a compressive strain, and the inner a tensile strain. 
When a hollow cylinder is cast, and is required to withstand a 
great bursting pressure — that is, all the metal is required to 
withstand tensile tresses — it is usual to cool it from the inside 
by means of a metal core, in which cold water circulates. The 
inside now becomes rigid sooner, the outer portions as they 
solidify contract, and tend to make the inner portion con- 
tract more than it naturally would, and there is a per- 
manent state of compressive strain inside and tensile strain 
outside in the object, which materially helps it to resist 
a bursting pressure. This inequality of contraction and 
production of internal strains in objects cause them to vary in 
their total bulk as compared with that of their patterns, but it 
is probable that some of this variation is due to the fact that 
the contraction of grey cast iron is only 1 per cent, of its 
linear dimension, whereas white cast iron contracts 2 to 
2 J per cent. The fractional difference between size of pattern 
and the finished object varies from one-twenty-fifth of an 
inch per foot in small, thin objects to one-eighth of an inch 
per foot in heavy pipe castings and girders. Small castings 
seem to be considerably stronger than large ones. As there is 
always great inequality in the rate of cooling of a casting near 
a sharp corner, internal strains may be expected here, and also 
an inequality in the nature of the cast iron, since the grey 
variety gets whiter the more rapidjy it is cooled. In nearly all 
bodies a re-entrant corner is a place of weakness (see Art. 303), 
and is specially to be guarded against in castings. Crystals of 
cast iron and other metals group themselves along lines of flow 
of heat. When a plate or wire of iron or steel is rolled or 
pulled, the crystals become more longitudinal, and the wire or 
plate becomes stronger, whereas annealing allows the crystal? 
to arrange themselves laterally, and the material is weakened 



284 APPLIED MECHANICS. 

It is said that time gradually reduces some internal strains. 
Castings which have been rapidly cooled by being cast in an 
iron mould (painted on its inside with loam) are ivhite, and 
very hard in those parts which lie nearest the mould, 
whereas they are grey and strong inside. These, like the 
hollow cylinder above mentioned, are called chilled castings. 
When a cleaned casting, preferably of white iron, is put in a 
box, surrounded witli oxide of iron (hematite iron ore or rolling 
mill scale), and kept at a white heat for a length of time (say 
a week), its surface, to a depth dependent on the time, loses 
its carbon and becomes pure or wrought iron, which is much 
tougher than cast iron. The teeth of wheels are sometimes 
heated in this way. Such are malleable castings. Malleable 
cast iron seems to be of 50 per cent, greater tensile strength than 
cast iron, with a contraction of 8 per cent, before fracture ; it 
stands about 60 tons per square inch crushing stress. Melted cast 
iron possesses the property of dissolving pieces of wrought iron, 
and is then said to be toughened cast iron. When sufficient 
iron is so added it becomes an inferior variety of cast steel. 

226. Wrought Iron.— Cast iron is exposed to the air in a 
melted state for a long time, and the carbon is burnt out of it. 
The pig-iron really undergoes two processes, one called refining^ 
the other puddling. It is then hammered and rolled, when hot, 
into bars of various shapes. The quality of wrought-iron bars 
as bought in the market varies greatly. We have common iron, 
used for rails, ships, and bridges ; best, double best, and treble 
best Staffordshire iron, used for boilers and forgings generally ; 
Lowmoor, Bowling, and other good irons for the most difficult 
forgings ; and, lastly, charcoal iron, which is nearly pure. Iron 
is softer and more ductile the purer it is. Traces of sulphur 
make it red-short, difficult to work hot. Phosphorus has 
effects like carbon, but also makes the iron cold-short, or 
treacherous eold. Maganese and silicon seem also to act like 
carbon, but we are still afraid of them. There is usually \ to 
J of 1' per cent, of manganese present, and one-third of these 
amounts of silicon if castings are wanted. Up to the tempera- 
tures of ordinary boilers, the tensile strength of iron is not much 
diminished by heating, but at a red heat it is very much less 
than in the cold state. Repeated forging increases the strength 
of wrought iron up to a certain number of times, after which 
it diminishes the strength. This is why small rods and small 
forgings and the outside of large forgings are respectively of 



APPLIED MECHANICS. 285 

stronger material than large rods and large forgings and the 
inside of large forgings. Cuts of metal in certain directions 
from heavy forgings seem surprisingly weak. By rolling and 
hammering when hot, iron gets a hbrous texture, and becomes 
more tenacious. By hammering when cold, or by long con- 
tinued strains of a vibratory kind, wrought iron changes its 
fibrous and tough for a crystallised and more brittle condition. 
This brittle condition may be removed by heating and slowly 
cooling (annealing). Iron wire is stronger the thinner it is. 
Bar iron is generally stronger than angle or T-iron, and this 
again than plate iron. The toughness of an iron bar is best 
shown by the contraction it undergoes before it breaks. The 
section of a very tough bar may contract as much as 45 per 
cent, in area. Case hardening of a wrought-iron object is 
effected by heating it in a box with bone dust and horn 
shavings. The iron absorbs carbon, and is partially converted 
into steel. 

227. Steel. — Steel contains less carbon and impurities than 
cast iron, and thus lies intermediatebet ween castironand wrought 
iron. Expensive steel is produced by giving carbon to wrought 
iron (the best Swedish charcoal iron), keeping the iron heated 
for some days in contact with powdered charcoal, and then 
hammering it, whilst hot, till it is homogeneous (shear steel), or 
else (and this is the most usual practice) casting it when melted 
into ingots. 

Cheap steel is produced by taking only a portion of the 
carbon from very pure varieties of cast iron by a puddling 
process such as is employed in the production of wrought iron, 
or by the Bessemer process. In the Bessemer process, air is 
forced into the melted cast iron for a time, and very pure white 
cast iron is then added to help in removing bubbles of gas. In 
Art. 235 I shall speak about the tempering of steel. Many 
varieties of soft steel do not harden when suddenly cooled. 
Some of it is like pure iron almost, and some of it has half as 
much carbon as the hardest cast steel. It differs from wrought 
iron in having been melted, and so there are no minute streaks 
of slag giving heterogeneousness ; and so a plate is nearly as 
strong one way as another. The sudden hardening of steel when 
rapidly cooled seems greater the more carbon there is, up to a 
certain limit. It is more fusible than wrought iron, and much 
success has been met with in the production of steel castings in 
spite of the fact that unless certain precautions are taken, and the 



286 



A.PPLIED MECHANICS. 



addition of silicon, aluminium, etc., there is a tendency to con- 
tain cavities. This steel has about twice the strength of cast 
iron. Annealing is necessary after casting. The strength of 
crucible cast steel is greater than that of any other material, and 
is greater as it contains more carbon and is harder. The pro- 
perties of steel depend so much on so many seemingly small 
things — small impurities, a little too much heating or variation 
in the rate of cooling at different places — that great care must 
be taken in working it. By the Bessemer, Basic, and Siemens 
processes great quantities of steel are produced cheaply, contain- 
ing small percentages of carbon. This steel has largely replaced 
wrought iron in its use in locomotive rails, bridges, and ships. 
Taking it that crucible steel has elastic limits of stress 26 
to 20 tons per square inch, and ultimate stress 52 to 34 tons, 
and contraction of section 5 to 20 per cent, before fracture, 
these are for Bessemer steel 17 tons, 34 tons, and 20 to 45 per 
cent. Steels to be easily welded together ought to be of the 
same kind. 

As steel cools from a welding heat it passes through the 
" blue heat " stage (about 300° C), where it is brittle. If 
hammered or bent in this state it is permanently iujured, and 
if the work was local in a large plate the plate becomes 
treacherous. Soft steel and iron seem to get a little stronger 
at very low temperatures. 

The amount of carbon present in steel does not seem to 
affect much the Young's modulus, which in all kinds of steel 
and wrought iron seems to be about 3 x 10^, as the modulus 
of rigidity is not very different from 12 x 10^. Various 
numbers are given in Table III., in deference to custom. The 
carbon produces other effects, shown in the following table : — 



% Carbon. 


Elastic limit 
stress. 


Breaking 

stress. 


Contraction of 
area. 


Ultimate 
shearing 

stress. 


•u 

•51 

•96 


18 
22 
31 


28 
36 
53 


•50 
•25 
•10 


22 
26 
37 



Stresses are in tons per square inch. A formula has been 
given :— Strengths: 19-5 + 11-4: C^ + 302 + 11-4 Mn -f 9-5 P, 
and elongation per cent. = 42 - 36 0. — 5-5 Mn - 6 Si, where 



APPLIED MECHANICS. 287 

C, Mn, P, and Si represent the fractional amounts of carbon, 
manganese, phosphorus, and silicon in the steel. 

The oxide and silicate skin on cast iron is less liable to 
corrosion than clean iron, but it is advisable to paint or varnish 
all iron exposed to oxidation. Sometimes, as for water-pipes, 
the iron is heated to 150° 0,, and placed in pitch with some oil 
in it at 100° C. Mr. Barff keeps the iron surface exposed to 
superheated steam at a high temperature, and this coats it with 
a film of protecting oxide. Iron is " galvanised " by putting 
it in a bat': of melted zinc. Boilers are sometimes protected 
inside by the contact of blocks of zinc. 

Some alloys of iron and manganese are very strong, and so 
hard as to prevent their being readily tooled. They can, how- 
ever, be both cast and forged. They are specially important to 
electrical engineers, as they are not magnetic. The power of a 
small trace of manganese to destroy all the magnetic properties 
of iron is remarkable. Certain alloys of iron and nickel are not 
quite so hard, but they are extremely tough and strong. Steels 
containing a little tungsten or chromium have the special 
properties that fit them for self-hardening tools or projectiles. 

Mitis castings are of wrought iron, to which 0*5 to 1 per 
cent, of aluminium has been added to lower the fusing point. 
The tenacity seems to be 20 per cent, greater than that of 
wrought iron, and the ductility about equal to that of wrought 
iron. 

228. Copper is noted for its malleability and ductility when 
both hot and cold, so that it is readily hammered into any shape, 
rolled into plates, and drawn into wires. When cast it usually 
contains much oxide and many cavities, but when pure it may 
be worked up by hammering into a state of great strength and 
toughness, whereas slight traces of carbon, sulphur, and other 
impurities necessitate its being refined to do away with its 
brittleness. The brittleness produced by hammering when cold 
is very different, as it is removable by annealing. Phosphorus 
is sometimes used to assist casting, and the strength is greater 
with more phosphorus, whose function seems that of reducing 
the oxides. Copper is an expensive metal, and is only used 
now for pipes which require to be bent cold, for bolts and 
plates in places where iron would be more readily corroded, 
and for electrical purposes. Its tensile strength is more 
reduced by heating than that of iron. 

Boron seems to affect copper as carbon does iron, the wire 



288 APPLIED MECHAXICS. 

alloy standing 22 to 27 tons per square inch vrithout loss of 
conductivity. 

229. Brass consists of about two parts by weight of copper to 
one of zinc, with a little tin and lead. The copper is first melted 
and the zinc added, not long before casting. It is used chiefly 
on account of its fine appearance and the ease with which it 
can be worked. Cheap brass things have more zinc usually. 
The small amount of lead added in melting makes it much 
softer, Muntz metal contains more zinc than ordinary brass — 
3 copper to 2 zinc, or 2 to 1 with a little lead. Like copper, it 
is not much weakened by heating up to 260" C. ; it can be 
rolled hot. It is used for sheathing ships and for the tubes 
of boilers. 

230. Sterro and Delta metal are brasses to which iron has 
been added, the latter name being given to the metal after it 
has also received special mechanical treatment, the rods being 
made by being forced to flow under great pressure through 
dies. Delta metal can be worked hot or cold, and may be 
brazed. Bronze and gun-metal are alloys of copper and tin in 
varying proportions, more tin giving greater hardness. Twelve 
of copper to 1 of tin seems best for guns. Five of copper to 
1 of tin is the hardest alloy used by the engineer in bear- 
ings, but bell metal is 3-3 to 1. A slight addition of zinc helps 
in casting, and increases the malleability. A great many 
experiments have been made on bronze. Its strength depends 
very much upon the care taken in mixing and melting the 
metals, for it is easily injure-d by oxidation. Gun-metal is a 
good material for castings, which, however, should be quickly 
cooled to give more uniformity, density, strength, and toughness ; 
and hence they are sometimes made in cast-iron moulds. Hard 
bronze is much used for the bearings of shafts. There are also 
various soft alloys of copper with lead, zinc, tin, and antimony, 
which are used for this purpose. Phosphor bronze is an alloy 
of copper and tin to which some phosj^horus has been added. 
It bears re-melting better than gun-metal, and its properties 
may be varied at will. It may be either strong and hard, or 
weaker but ^ery tough. The phosphorus appears to act by 
removal of the oxide of tin, and to tend to prevent segregation 
in cooling ; and, indeed, we may say that when care is taken 
against oxidation the difficulties in the way of re-melting gun- 
metal vanish. Hard wire has broken with from 100 to 150 
tons per square inch, and after annealing has stood half these 



APPLIED MECHANICS. 289 

amounts. It has been used successfully in railway axle and 
crank shaft bearings. It is good in resisting shocks, and has 
been used instead of steel for chisels in powder factories. 
Gun-metal slowly decreases in strength as it is heated, until at 
a certain temperature its strength is suddenly halved, and there 
is almost no ductility. Phosphor bronze is less affected. 

231. Manganese bronze and Silicon bronze are special alloys 
of copper with manganese and with silicon, the former made by 
adding ferro-manganese to bronze or brass, and used where 
unusual strength and power (as for screw propeller blades) of 
resisting sea-water are required. It may be cast and also 
forged. The manganese seems to act like phosphorus in 
clearing off" the oxide. Some containing zinc can be forged 
and rolled hot. Silicon bronze has fair electric conductivity, 
and resists atmospheric corrosion when used as telephone wire, 
and is of great strength. As tested for tension in the condi- 
tion of wire for telephonic purposes, it seems to stand from 30 
to 50 tons per square inch; the drawn copper wire for the 
same purpose standing about 29 tons to the square inch; 
phosphor bronze, 44 to 70 ; brass, 25 ; German silver, 30 ; 
iron, 57 ; Martin steel, 86 ; and crucible steel, 102 tons per 
square inch. Pianoforte wire sometimes stands 150 tons per 
square inch. 

232. Aluminium bronze is formed of 9 parts of copper and 
1 of aluminium, and has a tenacity of 43 tons per square inch. 
The 5 per cent, alloy stands 30 tons per square inch. Copper 
with only 2 to 3 per cent, of aluminium is stronger than brass. 
The usual alloy is of the colour of gold, but, like all aluminium 
alloys, must be prepared from materials free from iron. When 
ad\antage is to be taken of the lightness of aluminium, there 
is an alloy of 32 parts of aluminium and 1 of nickel that can 
be employed, and to which 1 part of copper may also be added. 
Specimens have broken with 45 and 50 tons per square inch 
tensile stress, the first with no elongation and the second with 
33 per cent, elongation. The aluminium and silicon bronzes 
and alloys of silver and other metals with aluminium are pro- 
duced in an electric furnace, the ores being mixed with retort 
carbon and an electric current passed. 

About 7 per cent, each of copper and antimony being 
added to tin, we have white metal or Babbitt's metal, which 
fuse easily and may be cast inside a bracket, round a journal, 
as the step of a bearing if not too large 



290 



CHAPTER XIII. 

TENSION AND COMPRESSION. 



233. The following chapter on the behaviour of material 
when subjected to tension and to compression has so much to 
do with the physical properties of matter, that students ought, 
before reading it, to refresh their memories in regard to the sim- 
pler principles of chemistry and physics and the notions which 
laboratory work in these subjects gives to us in regard to the 
probable molecular condition of matter, and also of that very 
much larger coarse-grainedness which we sometimes call hetero- 
geneity. Besides giving us general notions, chemistry gives us 
useful facts as to the changes which occur in the manufacture 
of metals, the cause of the rusting of metal, the burning of 
fuel, etc. A little knowledge of electricity enables us to have 
clear ideas as to the action by which when two metals touch, 
and are also connected by liquid, one of them rapidly corrodes 
and the other does not, and how it is that oil preserves a 
polished metal surface. A little knowledge of heat tells us 
how friction wastes mechanical energy ; how heat energy is 
measured ; when a body is heated how much it expands ; the 
laws of expansion of gases ; the properties of steam ; the laws 
of flow of heat in conduction and radiation, and other pheno- 
mena which continually influence our mere mechanical work. 

234. It will be found by students who read what is in the 
smaller printing in this chapter that the usual statements on 
which we base our mathematical calculations are very incom- 
plete. The manufacturer depends on many curious properties 
of materials, many of which are familiar to and helpful to 
inarticulate workmen and unknown in the laboratory as yet. 
Some rude processes and shop beliefs, which sometimes seem 
to be uo more worthy of attention than superstitions, have 
suggested scientific experiments and new industries. Some 
phenomena that seem curious at first sight are really easily ex- 
plained — the behaviour of James Thomson's overtwisted shaft, 
for example, explains many curious things ; and our knowledge 
of such things as how initial strains are induced in castings by 
unequal coolinghas helped us greatly in systematising our notions. 

235. There is one phenomenon which has been known .almost 



APPLIED MECHANICS. 291 

since iron was discovered by man — namely, that steel hardens 
with sudden cooling, and we seem to be almost as far from 
understanding it as our ancestors were. There has been some 
advance, for we have no such notions about incantations and 
the virtues of particular kinds of water for quenching the heat 
as were held in the Middle Ages. Mild steel is almost like pure 
iron, and does not harden, but with more carbon (over 0*4 per 
cent.), such as there is in cast steel, the more rapidly the steel 
is cooled the harder it gets. So that, for example, thin pieces 
of steel are apt to be harder than thick pieces. The more 
carbon a steel contains, the less need it be heated before being 
suddenly cooled to acquire a particular hardness. Re-heating 
diminishes the hardness, or, as it is called, tempers the steel ; 
and the higher the temperature of re-heating, the softer does 
the steel become. It may roughly be taken that sudden cool- 
ing in oil doubles the proof tensile strength of the material — 
that is, the stress which would permanently hurt the material. 

In every workshop the common method adopted for 
tempering a fitter's chisel is as follows : — Heat the chisel to 
a dull red colour, put the edge in water to a distance of say 
half an inch, so that it may become very hard; quickly brighten 
the edge with pumice or a file ; watch it till, as it heats by con- 
duction from the thicker portion, you know that a certain 
temperature has been reached by seeing a certain colour 
(purplish-yellow for a chisel) of oxide of iron making its 
appearance. When this colour appears, plunge the whole chisel 
into water. Thus the steel is first made extremely hard at its 
edge, and is then brought back to the required degree of 
hardness by re-heating up to a certain temperature and then 
cooling. This simple process is in common use. In tempering 
other objects sometimes much greater care must be taken, since 
it is often necessary that every portion of the object shall be of 
the same hardness, and in such cases the whole may be cooled 
at first and then re-heated in a bath of oil, mercury, or other 
melted metal whose temperature is definitely known. The 
effect is of the same kind, however, whether the process is the 
rough one which I have described or a more careful one. 

There must be no attempt to make large objects glass-hard ; 
they would cool very unequally and might fly to pieces or 
develop flaws ; a less rapid cooling in li/ot oil or melted lead 
tempers such objects in one process. It is interesting to see 
that one or other of the above two principles is carried out in 



292 APPLIED MFXHANICS. 

all sorts of iudustries, but in a great number of different ways. 
A certain size of watchinake:'s drill is stuck when at a red 
hec^.t into sealing-wax. This gives the right temper. Another 
smaller drill is merely waved about in the atmosphere to cool 
it. The tempering colours of steel, beginning with the hardest, 
are : — Straw colour to yellow, (this is about 220'Cj for light turn- 
ing tools, milling cutters, screw-cutting dies and taps, punches, 
chasers ; straw to purplish-yellow, rimers, wood-chisels, plane- 
irons, twist-drills ; light purple to dark blue, augers, chisels for 
steel, axes, chisels for cast iron, chisels for wrought iron, saws 
for metal ; less dark blue (this is about 320°C), screw-drivers 
and springs. These colours are supposed to indicate fairly 
exactly the temperature, irrespective of time ; but we cannot 
say that there is conclusive evidence yet that time produces no 
effect on the thickness of the film of oxide. If true, it is a very 
curious phenomenon. But surely it cannot be true ! 

236. The vohune gets greater in hardening. Curiously enoiigli> 
I have a note sa^-ingthat steel becomes denser hy hardening, but its 
authority is unknown. Eepcated hardening and annealing seem 
to strengthen the steel. It is usual to explain the hardemng "by 
saying that in sudden cooling the particles of iron and carbon have 
not had time to get into their natural positions when cold, and that 
they jam one another somehow, getting into positions of instabihty. 
As regards the influence of impurities, of gases from the atmosphere 
which may possibly be suddenly imprisoned among the particles, 
very Httle is known. It may help towards an explanation to say 
that Abel found that in annealed steel the carbon is in the form of 
a chemical carbide Fe.^C mixed in the mass. In hardening, the 
formation of the carbide is prevented (just as suddenly- cooled 
gases remain dissociated) . At various tempers we have various pro- 
portions of the carbide, but it is always the same kind of carbide. 

Speculation as to the molecular constitution of iron does not 
yet seem to have sufficient facts to go upon. It is sometimes 
assumed that there are two kinds of iron mixed together, the soft 
o particles and the harder B particles. "With slow cooHng from a 
high temperatucre, wben the mass is soft, although the particles 
may be hard, the /8 particles (practically all are of the /3 kind at a 
hi.o-b temperature) change to a. That there is some great molecular 
cl^nge even in the purest iron is evidenced by recalescence and 
other aUied phenomena. In sudden coohng most of the )8 particles 
have no time to change. Any effect due to the carbon is produced 
at a much lower temperature than that at which the change from 
/3 to a occurs in slow cooling ; and although the presence of carbon 
seems necessary to the hardening of steel, changes in its m^de of 
existence are not of much importance. The a particles are cbanged 
to 3 in the plastic condition of iron in the ordinary testing opera- 
tions at low temperatures. 



APPLIED MECHANICS. 



293 



237. How a pull is exerted. — How is it that a cord transmits 
force from my hand to an object when I pull the object by- 
means of a string ? If yoa study this matter, you will see that 
every particle of the string coheres to the next ; and although 
the refusal of one particle to come away from its neighbour 
might easily be overcome, there are so 

many of them to be separated at any 
particular section of the string that it 
requires a considerable pull to perform 
this operation. When a string is pulled 
it really lengthens a little, and it lengthens 
more the more force is applied, although 
it may not break. A string is not so easy 
to experiment w^ith as a wire of metal, 
because we find that it differs more in its 
quality at different sections, and it is 
affected by dampness and many other 
circumstances. No doubt it is also dif- 
ficult to obtain a metal wire which shall 
jugt be as willing to break at one place 
as another — that is, which shall be exactly 
of the same material everywhere ; but 
metal wire is certainly more uniform 
than string. 

238. Strain. — Take, then, a steel wire, 
A B (Fig. 170), fastened near the ceiling at 
A, between two pieces of wood, screwed 
together firmly so that there may be no 
tendency for the wire to break just at the 
fastening. Similarly fasten at b a scale- 
pan arrangement, and first place just so 
much weight in the pan as keeps the wire 
taut. Let there be two light little 
pointers stuck or tied on at a and h, and 
let there be a vertical scale on the wall. 
Now read off the distance between a and h on the scale, and 
note the weight. Add more weight, and again read the 
distance, and continue doing this until the wire breaks. You 
will prove by means of squared paper that the amount of the 
extension of a wire is nearly proportional to the weight which 
produces the extension. 

239. My students are in the hahit of using a more exact method 




I'ig. 170. 



294 



APPLIED MECHANICS. 



of measurement, a scale hanging from a and two verniers being 
attached to the wire at a and b. They also use a method in which 
a vernier on a is hrought close to a scale on h, the wire heing 
passed over pulleys. In some cases they use micrometer methods 
of measurement for greater accuracy, and they also experiment 
with larger specimens, loading them hy means of levers or wheels 
and screws ; and advanced students may he allowed to use 
machines in which loads up to 100 tons are applied, and arrange- 
ments may be employed for making automatic records of the load 
and the extension. If, however, the apparatus is elaborate and 
imposing as compared with the specimen, a beginner cannot 
readily pick up the essential idea of an experiment, and hence he 
had better begin with visible specimens loaded with visible weights'. 
He n.ay proceed to use such a machine as Bailey's wire-testing 

machine, and after- 
wards make a few 
tests with large 
commercial testing- 
machines. 

Bailey's latest 
form of machine is 
shown in Fig. 171. 
The specimen, say 
of l^-inch wire, is 
shown at d, being 
grij)ped at c and b. 
By turning the 
handle, a, we turn 
a worm driving a 
worm-wheel, turn- 
ing a screw whose 
nut is part of the 
frame, and so the 
gripping piece b pulls on the specimen d. The other gripping 
piece, c, tilts the weight, r, and the amount of tilting which measm-es 
the pulling force is indicated by a pointer on the dial, e. 

In some English engineering laboratories experimenting with a 
steam-engine and testing specimens in tension by means of a large 
testing-machine are sujDposed to be the only experimental exercises 
in which students ought to engage. Tests of specimens in com- 
pression, bending, and twisting are, however, sometimes made. 
Consequently a description of all the kinds of large testing-machines 
which have ever been constructed forms a large part of the college 
instruction in mechanical engineering. It seems to me, however, 
somewhat out of place in a text-book, as almost every student has 
opportunities of examining some such machine for himself. Complete 
information will be found in a paper b)^ Prof. Kennedy read before 
the Inst, of Civ. Eng. (Proceedings I.C.E., vol. Ixxxviii., 1889). 

The machines in most common use apply tensile load to the 
lower end of a test-piece by means of an hydraulic press. The 
upper end is pulled by means of a lever (whose fulcrum is a Imife- 
edge), over which a weight may be rolled by machii.ery into such 




Fig. 170a. 



APPLIED MECHANICS. 



296 




Fig. 171. 



positions tliat the lever is kept horizontal ; the position of the 
weight measures the pull. 

Instruments have heen designed which register on a sheet of 
paper (as the pencil of a steam-engine indicator does) the load 
pulling a rod, and the extension which it produces. A little brass 
cylinder covered with paper is touched by a pencil on tbe end 
of the rod. The amount of rotation of the barrel is regulated 
so that it is proportional to the load. By this means cm-ves like 
that of Fig. 172 may rapidly be drawn as the load on the rod is 
gradually made to increase till the rod breaks (see Art. 244). 

When Young's modulus for a material is wanted, my advanced 
students employ Prof. Ewing's extensometer. A half-inch bar of 
iron being given, even a load of 40 lus. causes sufficient extension 



296 APPLIED MECHANICS. 

of the distance between two marks about a foot asunder to be 
measured with sufficient accuracy for the determination of Young-'e 
modulus by tliis beautiful instrument, shown in Fig. 170a. 

240. When we speak of the tensile strain in the wire, and 
ivant to use the term strain in an exact sense, we mean the 

fraction of itself by which a b lengthens. Thus, suppose that 
a b was 50 feet and that it lengthens 1 foot, we say that the 
strain is ^\, or -02, or 2 per cent. I need hardly tell you 
how important it is to learn the exact meaning of a word 
like this. 

241. Stress. — If you take another wire of the same material, 
but of twice the sectional area of this one, you will find that 
it needs twice as much load to produce the same strain. The 
reason of this is somehow due to the fact that you have at any 
section twice as many particles of steel resisting the pull. The 
pull produced by the load acts at every cross-section in the 
same way, no matter how long the wire may be ; * but if the 
wire is thicker at one place than another, then at such a cross- 
section the pull is distributed over a greater number of pairs 
of particles. We see, then, that if a wire or rod is transmit- 
ting a pull, it is well not to consider the total load, but rather 
the load per square inch of section. The load per square inch is 
called the stress. 

* One important result of St. Veiiant's investigation (see Art. 306) is that 
the actual distribution of the load on a small area is not important. 
At a point not very near, the strain will be the same whatever the distribu- 
tion of load. Near the gripping-places and places of rapid changes of section 
in om' specimens, the stress cannot be expected to be uniform throughout a 
cross-section ; hence test-pieces are made larger near the grips, as we do not 
wish to break or study the specimen at a place where the distribution of 
stress is unknown to us. This is particularly noticeable at a narrow neck cut 
in a round specimen. There is great stress near the ends of tbe neck, 
inducing fracture there more readily than elsewhere if the material is hard, 
but in more ductile material inducing flowing of the metal, which prevents 
the section of the neck becoming as small before fracture as a long specimen 
wovild become, and so increasing the apparent strength of the material. This 
causes a strip of boiler-plate with a drilled or punched hole (tbe plate must 
be annealed after punching to destroy local hardness) to seem stronger per 
square inch of material left at the sides of the hole. If a specimen is too 
short, it does not seem to have freedom to contract as much in section before 
fracture, and therefore the breaking stress is greater and the fractional 
elongation is less ; for it is to be remembered that breaking stress is 
calculated as breaking load per square inch of the original section. Very 
long rods or wires give too small a breaking stress for a very different reason 
—namely, because of the greater chance of the existence somewhere of 
inferior metal. It is reasonable to suppose, and it is found experimentally to 
be fairly correct, that to obtain the same fractional elongations from the same 
material, specimens, if of different sizes, ought to be similar. 



APPLIED MECHANICS. 29? 

This is the exact meaning which we give to the word 
stress. Much of the difficulty you may have met with in your 
reading is due to the fact that you have not made a proper 
distinction between the meanings of these two words. Stress 
is the load per square inch which produces a fractional altera- 
tion of the length of a wire or rod, and this fractional alteration 
is called the strain. Suppose your load to be 6 lbs., and your 
wire circular in section, with a diameter of 0'05 inch. Then 
the area of the section is 0-025 x 0025 x 3-1416, or -00196 
square inch. The stress is 6 -f- -00196, or 3,061 lbs. per 
square inch. You will find that this thin wire gets the same 
strain with a total load of 6 lbs. as a rod 1 square inch in 
section would get with a load of 3,061 lbs. If ever you get a 
problem to work out relating to the lengthening of a wire or 
rod produced by a load, you must consider not the total 
lengthening of the wire or rod, but its fractional amount of 
lengthening, and call this the strain; also consider not the 
total load, but the load per square inch of section, and call 
this the stress, and you will find that for some kinds of 
wrought iron the tensile stress = the tensile strain x 29,000,000. 

242. It is iisixally assiuned that we ougM to expect the elongation 
and strength to depend upon load jper square inch and not njDon 
the shape of the section. To me it seems wonderful that mole- 
cules near the surface should not behave differently from molecrdes 
remote from the surface. It is possible that the mere shape may 
have some small effect which is disguised for us by the great 
differences of material and of physical state in the specimens 
which we compare. It is only when we take pains to obtain 
homogeneous material that we find the strain very nearly propor- 
tional to the stress. Any departure from this rule may be traced 
to initial strains in the specimens. Unless a casting or cold- 
hammered or swaged forging is annealed by heating to a bright 
red heat and a slow cooling, it is heterogeneous. Some parts take 
a set much more easily than others, because they are of different 
material or because they are already strained. Instead of Fig. 172, 
therefore, we ought to expect a figure which is the result of adding 
together the ordinates of a great number of curves like Fig. 172, 
the distances o n being very different, in some cases 0. A cast- 
iron beam takes a set for quite small loads, and it is often loaded 
so as to get a large set before it leaves the foundry. Afterwards 
its deflection will be jDractically jDroportional to load. A man who 
puts up bells in a house or a telegraph wire " kills " the wii-e 
beforehand — that is, gives it a pennanent set, straining it to its 
"yield point" indeed, as he finds that after this operation it is 
harder, wall stand great loads without taking a further set, and 
follows the laws of elasticity better for pulling forces. This is like 
giving a good set to a piece of riveted work, which means that the 



298 AI'PLIED MECHA^^lCg. 

rivets bed better into tbeir boles. Wrhen a wire by being draWH 
through a die is reduced to a smaller size, there is a complete 
alteration in the arrangement of its particles or groups of 
molecules, and yet the drawn wii-e has usually greater strength 
than it had originally. Even hardened steel wire is drawn in this 
way. Metals will, in fact, flow (the filling up of the passages in 
mines shows that rocks also flow) - if sufficient stress is applied to 
them, and at the end of the operation they are as strong or stronger, 
but less plastic, than before. If they are harder and this quality 
is not wanted, it may usually be removed by annealing. Plates 
of iron and steel rolled cold are hardened ; rolled hot, they are 
gradually annealed as they leave the rolls. In constructing a 
certain magnetic instrument, I find it necessary to anneal a certain 
piece of iron from so high a temperature that the nearly pure iron 
is so soft that it almost cannot keej) in shape. If this is scratched 
once by a file, sufficient hardness is induced to make the instrument 
useless. 

243. It is somewhat more difficult to experiment on the 
shortening of a strut or column when it transmits a push, 
because you cannot use very long struts. A strut tends to bend 
if it is very long ; and when it breaks, unless great care is 
taken to keep it straight, it breaks more easily the longer it is. 
The bending action causes the load to act more on one part of 
the cross-section than another, and the stress — or the pushing 
force per square inch — is greater at one part of the section 
than at another. If you experiment, therefore, you must take 
care to use struts which are prevented from bending. In 
Chap. XYI. we shall consider the bending of beams, after which 
you will better understand the present difficulty. It is sufficient 
for 3'ou at present to know that whereas the pull in a tie-bar 
tends to make it straighter if possible, the push in a strut tends 
to make it bend. Hence, in an iron railway-bridge or roof 
you will see that the tie-bars are thin solid rods usually, and they 
might be chains or ropes ; but the struts must not merely have 
a proper area of cross-section, this cross-section must also be 
wide in every direction. Thus, instead of a solid cast-iron 
column you always see a hollow one, unless the column is very 
short. Also, a thin plate of iron suffices for the lower boom 
or flange of a railway-girder (because it resists a pull), whereas 

the top boom is a hollow tube, or is U-, or p|-, or i i-shaped, 

because it must resist a push. Long struts, therefore, must be 
considered in Chap. XXI., after we have investigated the bend- 
ing of beams. 

If, however, we prevent bending, the laws for stiffness and 
sueiigtli of long struts are as simple as those of tie-bars. Thus 



APPLIED MECHANICS. 299 

the great struts of the Forth Bridge consist merely of four angle 
irons arranged as the four parallel edges of a square prism, and 
they are fastened to one another laterally by very light bracing, 
whose simple function is to prevent the bending of the angle 
irons. The strength and stiffness of such a strut are to be 
calculated from the combined sections of the angle irons. 

In a short strut or a long strut prevented from bending (in 
Art. 373 it will be shown that the lateral constraints are called 
upon to exert only very small forces to prevent bending), the 
load per square inch is called the stress. The shortening is a 
fraction of the whole length of the strut, and this fraction is 
called the strain. You will find from your experiments that 
the strain is proportional to the stress. Thus for wrought-iron 
struts or columns the compressive stress = the compressive 
strain x 29,000,000. The multiplying number is found to be 
the same for the same material, whether it resists a push or a 
pull. This number is called " Young's Modulus of Elasticity " ; 
it has been measured for various materials, and is given in 
Table III., p. 302. In using it you must remember that the 
stress is in pounds per square inch. 

Exercise 1. — By how much would a round bar of steel, 120 
feet long, whose diameter is 2 inches, lengthen with a pull of 
30 tons 1 Answer :— 0-0855 foot. 

Exercise 2. — By how much would a column of oak, 7 feet 
long and 4 inches square, be compressed in supporting a weight 
of 2 tons 1 Answer :— 0013 foot. 

244. I have said that if you use squared paper after making 
your experiments, you will find that the strain is proportional 
to the stress, and the lengthening of a tie bar is proportional 
to the total pulling force. But you will find that this law is 
not true when the loads become too great. If your loads are 
less than about a quarter of the breaking load, you will find on 
removing them that the wire on which you are experim.enting 
goes back to its original length.* But if your loads much 

* It may not go back at once to its old length, but in a few minutes it 
will be found exactly where it was before you loaded it. Similarly, when 
the load is put on there is first a sudden lengthening, and after this there is a 
slight extension going on so long as the load remains, but it practically comes 
to an end in a few minutes. This after-action, or " creeping, " is so slight that I 
have not till now spoken about it, although we have reason to believe that its 
investigation would be of great importance. This "creeping," which seems 
connected with internal friction or viscosity, is absent in the quartz fibres of 
Professor Boy? within a large range of stress, and it is so in soft metals within 
a smaU range ; and this is cvirious, because there is much of it in glass ; and 



300 



APPLlEt) MECHANICS. 



exceed this amount it will be found that the wire has taken a 
permanent set : that is, if you remove the load the wire will 
not go back to its original length. It remains permanently 
longer than it originally was, and we say that we have exceeded 
the limits of elasticity.. The load which produces this per- 
manent set is said to be the measure of the elastic strength of 
the wire, for although it does not break the wire it alters it 
permanently. Now it is only for loads less than this that the 
law " strain is proportional to stress " is true. Your squared 
paper for experiments on a steel wire would give a straight line 
becoming a curve, like Fig. 172. 

When you plot your results, making the distance m n 
1 epresent the extension of the wire 
for a load represented by the dis- 
tance m q, to any scale you please, 
you will find that the line passing 
through your points is straight 
only from o to M, say, and then 
it curves. The distance Q m repre- 
sents the load which produces 
permanent set. For greater loads 
than this, the extension is more 
than proportional to the load, and 
increases more rapidly until we 
get at P K a very rapid extension 
indeed, and the elongations there- 
after are so great that they cannot 
be represented here on the same 
scale as the part o M. p is called 
the yield point, and the pheno- 
menon is very marked m wrought iron and other ductile 
materials. It seems always somewhat higher than M, the true 
elastic limit. A student must obtain results up to the breaking 

certainly within tlie limits of permanent set there is creeping in brittle sub- 
stances, such as steel, giving trouble in measuring-instruments. If we say that 
there is perfect elasticity when the same forces are permanently required to 
keep a body in any particular shape at the same temperature, a body may be 
perfectly elastic, and yet it may have viscosity (the term viscosity indicates 
an elasticity which is affected by the rate of change of strain), and disobey 
Hooke's law. But there is imperfect elasticity if the forces are dependent 
not only on the shape of the body, but on its previous shapes, its past history. 
The only perfectly definite limits of elasticity in nature seem to be those at 
which, at constant temperature, a vapour becomes a liquid and a liquid 
becomes a solid, or a vapour becomes a solid. 




q, Q 



Fig. 172. 



ked 



APPLIED MHCHANICS. 301 

of a wire for himself, plotting them on squared paper ; noting 
the rapid yield after p, then the much slower but still rapid 
elongation, becoming again rapid as the specimen gets near 
breaking. The material is plastic after m. He will note that 
published diagrams are not of great value unless we know the 
rate at which the load was increased. After p is reached the 
specimen will continue to increase in length with time, even when 
the load is not increased, and even when the load is diminished. 

245. These gi'eat plastic elongations are permanent (except 
for the very small elastic part), and are very different from the 
great non-permanent compression of cork, or the great extension or 
shortening with nearly constant volume of indiarubber. "WTien there 
is continuous yielding under constant forces, the solid body behaves 
like a fluid, flowing (as lead does easily) without much change 
of density. It seems from M. Tresca's experiments that when a 
round punch comes down upon a plate of lead the lead flows 
rapidly laterally from underneath the punch into the rest of the 
plate, until the shearing surface is considerably diminished, because 
the wad is much less thick than the plate from which it is 
punched. In the squeezing of metals there is local flow wherever 
the pressure is great. 

246. A fluid is defined as that which greatly changes its shape 
under the action of indefinitely small forces. This is why we call 
many substances (such as sealing-wax and pitch) fluids. Even 
their own weights, acting long enough, cause them to flow. Metal 
statues and brackets several thousands of years old prove that metals 
are not fluids. Experimental evidence of what stresses infinitely 
long continued will just cause metals to flow is still wanting. Mr. 
Bottomley found that seemingly similar wires seemed to greatly 
increase in their strength if the increase of load was very gradual 
indeed. The increase continuing in one case for a month, the 
strength seemed to be greater by 27 per cent, than on tbe specimen 
broken in the ordinary way. 

247. When after the yield-point, aload is left upon a specimen, the 
amount of hardness and increase of strength produced are increased 
by lea^dng on the load for a longer time. Even when the load is 
not left on, if the specimen is left unloaded for some time, it seems 
to become harder and considerably stronger during the interval of 
rest. In both these cases by hardness I mean a high yield-point. 
As a matter of fact, the Young's modulus seems to diminish 4 to 9 
per cent, as it gets what I now call harder, and annealing increases 
it again. _ The material below the ;^deld-point, although seeming 
very elastic, exhibits time plasticity. The nature of the change in 
the character of these tensile and compressive phenomena when 
other stresses (such as twistmg stresses) are acting is not yet weU 
known ; but there may be an apparent alteration in these limits, 
as deduced from the bending of beams or the twisting of shafts 
which have initial strains. Thus James Thomson showed that a 
shaft of ductile material might be twisted until most of the 



302 



APPLIED MECHANICS. 



material had taken a set. The |haft, if now examined, would be 
found to take a set very much more readily with the reversed kind 
of twisting couple than with one of the kind which was used to 
give it its set. To imderstand this properly, a student ought to 
make a diagram showing the prohahle stress everywhere in the 
shaft when left to itself. 

248. Moduli of elasticity are slightly altered by manufacturing 
processes. Lord Kelvin found copper and ii'on wise to alter 5 per 
cent, in rigidity when subjected to great stretching. Very much 
greater changes were produced in the Young's moduli by excessive 
twisting. Lord Kelvin found that good copper wire, annealed by 
being heated to redness and sudden cooling in cold water, had a 
modulus of rigidity (j-71 x 10*^ lbs. per square inch. The same kind 
of wire heated to redness and cooled slowly, so that it was brittle, 
had a modulus 5-58 x 10^, its density having diminished 2f percent. 

It is interesting to note that a watch goes faster and faster for 
some time after it is made, but at the end of some months the 
balance- spring settles down into a state which does not much 
change afterwards. In this state, then, its elasticity is greater 
than it was in the beginning. The springs of chronometers are, 
however, often laid aside as useless after a few years' ser'sdce, their 
elastic condition having altered so much since the beginning that 
they have to be replaced. ^ 

Young's modulus sometime^ diminishes and sometimes increases 
with temperature ; but more experiments are needed. Wertheim's 
results are given here : — 

TABLE III. 









Young's Modulus in Millions of lbs. 








per Square Inch. 




Metal. 


Specific Gravity. 










At 15° C. 


At 100° C. 


At 200° C. 


Lead 




11-232 


2-46 


2-32 




Gold 




18-035 


7-9 


7-5 


7-8 


Silver 


. 


10-304 


10-1 


10-3 


9-0 


Palladium 




11-225 


13-9 






Copper 





8-936 


14-9 


13-3 


11-2 


Platinum . 




21-083 


22-0 


20-1 


18-4 


Steel, Drawn, English ... 


7-622 


24-6 


30-3 


27-3 


Cast Steel 




7-919 


27-7 


27-0 


25-4 


Iron 





7-757 


29-6 


31-1 


25-2 



I cannot quite believe these results, because I have never 
known any specimen of steel to have a less e than 28 x 10^. They 
are reduced from the numbers as selected by liOrd Kelvin from 
Wertheim's original "Memoires"; and I must say that Wertheim's 
experiments were carefully canned out. 



APPLIED MECHANICS. 303 

The modulus of rigidity of iron, copper, and brass, according to 
Kohlraiisch, diminishes with temperature. The cubic modulus of 
water increases 15 per cent, for 53° C. rise of temperature; of 
alcohol and ether, it diminishes with temperature. The Young's 
modulus of indiarubber increases with temperature. The con- 
clusions which may be drawn by thermodynamic reasoning from 
facts like these are interesting. 

249. Some metals are beheved to become brittle at low tempera- 
tures, much as "cold-short" ii'on or steel containing j)hosphorus is 
brittle. In many cases there may be no proper foundation for this be- 
lief. There is a slight increase of strength in ordinary iron and steel 
.from low temperatures to near 200° C. Above about 300° 0. there 
is a great lowering of strength produced by rise of temperature. 
It is believed that iron and mild steel worked at a "blue heat" 
(lower than red heat) become more deteriorated (brittle, or with a 
tendency to become brittle afterwards) than if worked cold or at a 
red heat. 

250. Loss of Energy in Change of Strain. — ^Even in fluids 
there must be loss of thermodynamic energy in change of volume, 
because of change of temj)erature. But the loss of energy in the 
change of shape of solids depending on the rapidity of change, is 
much greater than can be accounted for thermodynamically or by 
motions of the outside air ; and it must be put down to internal 
fiiction or viscosity. It is very marked in zinc, in indiarubber, 
and in jeUies. The reasons for the viscosity shown in the 
distortion of gases and liquids are known to lis. In twistiiag and 
•untwisting wires. Lord Kelvin found (1) that the viscosity does not 
seem to be proportional to the velocity, so that the law is different 
from that of the viscosity of fluids. (2) Teiioion in the wire 
produces a non-permanent increase in the viscosity. (3) The 
viscosity at any time depends upon the history of the specimen. 
The -sdscosity of a wire on Saturday, after a week's experimenting, 
was greater than when on Monday the experiments were resumed 
after a Sunday's rest. Yoimg's modulus in iron and copper 
generally diminishes (in bronze it iucreases) with repetition of 
loading if there is little pause. The effect is not so marked if 
there is a pause. 

251. Strength. — Table XXII., p. 658, shows, among other 
things, the pulling (or tensile) and pushing (or compressive) 
stress which a material will bear before breaking. Probably if 
these stresses were allowed to act on the material for some 
time it would break even if they were not added to. They are 
obtained from experiments in which the load was increased 
pretty quickly, and yet quietly — that is, without any jerking or 
sudden action. The numbers in the table are taken from many 
sources, and must in general only be regarded as giving rough 
average values. Ultimate crushing stress is sometimes badly 
defined, because the materials behave as if gradually becoming 
plastic. Rolled iron gets a fibrous structure, and may not have 



304 APPLIED MECHANICS. 

the same strengths in all directions. Rolled steel is morft 
nniform. The smallest stress per square inch ivhich ivill 
produce a perinanent set in the maierial is sometimes called the 
elastic strength. The working stress is usually a fraction of 
this ; it is the stress which experience tells us to calculate 
upon for loads acting for a long time on materials, and which 
we shall be sure are perfectly safe in the case of such 
materials as are supplied from foundries and forges. 

Exercise 1. — How great a pull will a round rod of brass 
stand before it breaks, if its diameter is 0-3 inch"? What pull 
would produce in it a permanent set, and what is the safe 
working pull "l Answers : — 1,237 ; 484 ; and 254 lbs. 

Exercise 2.— A short hollow cylindric column of cast iron 
is 8 inches in outer diameter, 5 inches inner diameter. What 
is the safe load and what load will produce permanent set ? 
Answer: — The area of cross-section is 4 x 4 x 3"1416 minus 
2-5 X 2-5 X 3-1416, or 30-63 square inches; 30-63 x 21,000 is 
643,230 lbs., or 287 tons; 30-63 x 10,400 is 318,522 lbs., or 
142 tons. 

252. In making calculations on the stiffness and strength of 
structures we assume that strains are proportional to stresses. 
To use in these calculations the results of tests on materials 
beyond the elastic limit may be unscientific, but this is what 
we do : employing a factor of safety, which is the ultimate 
strength divided by the working strength. That is, we say 
that we may use a certain working load on a beam because we 
find the pushing or pulling forces which it produces in the 
various parts of the structure ; and each of these is, say, one 
quarter of the pushing or pulling force which would break that 
particular bar ; here the factor of safety is four. We do not 
mean, or, rather, we ought not to mean, that we might put four 
times our working load upon the structure before breaking, 
because we have no theory of what would occur in the struct".re 
if the elastic limit of one or more bars were passed. The factor 
of safety allows also for contingencies ; there is the risk of 
defects in spite of precautions and the possible deterioration with 
time; sometimes a large factor is used because we suspect 
inaccuracy in our theory or in our estimated loads, in unfore- 
seen causes of shock and fatigue. It will be evident from this 
and from what follows that great judgment is needed in fixing 
a factor of safety, and the following list must be looked upon 
merely as a guide in setting academic problems : — 



APPLIED MECHANICS. 
TABLE IV. 



305 





Steady Load. 


Varying Load. 


Structures Subjected 
to Shocks. 


Wrought Iron and Steel 

Cast Iron 

Timber 

Brickwork and Masoniy 


f 

7 
20 


5 to 8 

6 to 10 
10 to 15 

30 


10 to 12 
15 
20 



Live loads are usually doubled and added to the dead loads. 
These numbers are sometimes called "factors of our ignorance. " 
The Board of Trade in 1858 adopted the rule that in bridges 
the stress on wrought iron must not exceed 5 tons per square 
inch, and later for steel 6J tons per square inch. We know 
now from Wohler's experiments that the range of stress is very 
important. In the Conway bridge, in which the range is not 
great because the live load is small compared with the dead load, 
the stresses are as much as 6 tons to the square inch. 

253. The load per square inch on a tie or strut is the stress. It 
is obvious that as the section of a tie-bar becomes smaller, the stress 
is really the load per square inch of the diminished section. The 
?hange of area A section in our ordinary constructive materials is 
so little that we usually consider the stress to be load -~ original 
area of section, as this is very convenient. But the section of a 
bar may alter very greatly when being fractured ; and although 
no experimenter yet seems to have stated it, I am sure that a great 
deal of useful information is lost because we do not plot tensile 
strain and actual tensile stress up to the breaking-point. It is 
somewhat difficult to measure the section with gTeat accuracy, and 
indeed within the elastic limits the measurement is nearly im- 
possible; but it ought to be attempted above the yield-point, 
because the section alters sometimes 50 per cent, before fracture. 
The stress seems to increase very much before fracture. 

In indiarubber the alterations in shape with good elasticity are 
so very great in comparison with what we find in ordinary 
substances that it is absolutely necessary to measure the real 
stress for each load and length. Students ought to do this for 
themselves, as observations even roughly made give instructive 
results. It is to be remembered that in gases the cubical increase 

of stress 5jj produces the cubical strain ; and as elasticity = 



dp 



The actual 



stress -^ strain, the elasticity of a gas is — ^^^ . 

volume at the time is to be taken. If we define Young's modulus 



306 APPLIED .MECHANICS. 

dp 

load divided by the actual cross-section, we find that for india- 
rubber it is fairly constant for loads that shorten the specimen 16 
per cent, or lengthen it 50 per cent. The most striking- thing is 
the rather rapid increase of the resistance to elongation in the 
attenuated si^ecimen before it breaks. 

254. In some simple structm-es the value of a material is taken to 
depend upon the amount of change of shape which it may undergo 
before it breaks ; and commercial tests often consist of definite 
extreme bending or twisting, or the delivery of definite blows to 
definite specimens. 

255. At first the contraction of section occurs pretty unifonnly 
along the specimen, but as fracture approaches, the contraction (ex- 
cept in hard metals, such as cast iron and hard steel) becomes 
localised. Hence this contraction, and not the lengthening of the 
whole specimen, ought to be taken as a measure of the ductility. 
"When the stretching proceeds slowly, and there are intervals of 
rest, there is greater imiformity, places of great yielding harden, 
and the jdeld takes place afterwards elsewhere. But, again, there 
have been cases where the loading was so very rapid that almost 
no local excess of yielding occuiTcd, and the whole specimen 
yielded fairly evenly, the sjDecimen extending about 50 or 60 per 
cent., or about double what occurred in a similar specimen tested 
in the ordinary way. 

256. It is interesting to examine the surface of the fracture, 
noting its texture. The fracture surface may be a flat or an oblique 
cross-section (see Art. 290). In ductile material we often find a 
combination of the two kinds, sometimes taking the form of conical 
and flat surfaces. In compression in short specimens the usual 
way of applying load prevents by friction the lateral enlargement 
of the ends. The material takes a barrel shape, and can withstand 
enormous loads, because the internal strain is greatly of the nature 
of cubical strain, and it is only shear strain that seems to cause 
fractiu'e (see Art. 291). 

257. Art. 290 shows that the shear stress in a tie-bar or strut is 
greatest on a plane making 45 degTces with the axis. But we do 
not find that oblique sections of fracture are inclined at this angle, 
possibly for the reason there stated. Tensile stress seems to 
diminish and compressive stress to increase the resistance to shear- 
ing. When, as in blocks of cast iron which are not more than 
half as high again as the width of base, there is no room for one 
shearing surface at the angle preferred by the material, the block 
fractui'es at a number of surfaces, cones and wedges being formed. 

258. Attention ought to be paid to the fact that local strengthen- 
ing and stiffening of a structure may produce general weakness. 
When a hole is punched in an iron plate there is a local hardening, 
and probably strength-increase in the material round the hole ; but 
unless the plate is annealed or (as a partial improvement) the hole 
is rhymered out to remove the hard part, when load is applied the 
hard part yields less readily than the rest, gets an imdue share of 
load, fiactui'es, and then the softer material fiactures. 



APPLIED MECHANICS. 307 

259. Live Loads. — When a weight is suddenly applied to 
stretch a wire, it produces greater effects than when slowly and 
quietly applied. We know the reason. A weight which, slowly 
applied, would produce an extension of 1 inch, will, when put 
on and let go, produce an extension of 2 inches. The wire 
now shortens to its original length, then extends 2 inches, and 
continues to get shorter and longer, the weight vibrating. As 
there is friction of some kind among the particles of the wire, 
and there is also external friction, the lengthenings and 
shortenings gradually lessen till, in a short time, the wire 
settles down into the same state as it would have been in if 
the load had been slowly applied. Now, if we suppose this 
wire, when stretched 2 inches, to be strained just beyond its 
elastic strength, it is evident that the suddenly applied load 
does harm ; whereas tlie same load slowly applied woiJd do 
no harm. The harm is greater if the weight, besides being 
applied suddenly, is moving before it begins to act on the wire. 
Take the case of a stone which is being removed by means of a 
crane. If the stone, happening to fall a little, be brought up 
by the chain, the increase in the stress on the chain depends 
on the height from which the stone has fallen, and is 
greater the less the chain is extended. When a wire is 
lengthened '1 foot by a weight of 1,000 lbs., which has been 
increased gradually, we know that the pull on the wire began 
with 0, and, as the wire gradually extended, the pull became 
greater, till it is now 1,000 lbs. The average pull was 500 lbs., 
and 500 x -1, or 50 foot-pounds, is the total strain energy 
stored up in the wire. If we wish to give more energy to the 
wire, we must strain it more ; and this is just what we do 
when we let the weight fall suddenly. The extra strains due 
to loads being live depend upon the mass which we set in 
motion in applying the load. In some railway bridges it has 
been found that the increased deflection is 14 per cent, greater, 
and it is usual to add 50 or even 75 per cent, to such live 
loads and treat them as dead. 

260. The energy stored up in any strained body may be cal- 
culated if we know the stress and the strain. The mainspring- of a 
watch contains a store of energy which is gradually given out by 
the spring in returning to an unstrained condition. Each strained 
portion of the spring contains a portion of the store, and if at any- 
place in the body there is too great a store the body will break 
there. 

If w is the proof load (or load which will just fall short of 



308 APPLIED MECHANICS. 

producing permanent set) on a tie-bar of length I and cross-section 
A, the stress being w/a, the strain is w/ae, the elongation is wt'/AE, 
and the work done as the load is increased from to "w is ^w-^/ae. 
This is called the resilience of the bar. Xow, the Yolume of the 
bar is Ia, and hence the resilience per unit volume is -i-w^/A-E. 
Replacing w/a by/, the proof stress which the material wiU stand, 
we have the resilience per cubic inch to be /V^e. If/ is 
the proof stress in compression, we have the same expression 
for struts. 

When a torque m produces an angular change S0, the work 
done is m . 50 ; and when a twisting moment m has been gradually 
increased from to m, the twist of a shaft increasing from to d, 
the work done is \-iiQ. 

When there is a shear stress / and a corresponding shear strain 
//n, the shear strain energy per unit volume is /2/2x ; and if / is 
the proof shear stress, /2/2n is called the resilience of the material 
per unit volume. 

The resilience is the strain energy which material may store 
before permanent set takes place. It is evidently /2 -^ 2e per 
cubic inch in tie-bars and struts, if / is the tensile or compressive 
stress which v/ould produce permanent set and e is Young's 
modulus. In shear the resilience is /^ 4- 2n, if / is the limiting 
elastic shear stress and n the modulus of rigidity. When the stress 
is not uniform, as in beams and shafts, the average resilience per 
cubic inch is, of course, less. Shocks due to blows, as of falling 
weights, will often cause the strain energy to exceed the amount 
which the material will stand, and local set and plastic yielding 
may take place. Much depends upon the rate at which straia 
energy is carried off to the rest of the material. 

261. Let us consider why a chisel cuts into an iron plate. WTien 
I strike the head of a chisel with a hammer I give to the chisel in a 
very short period of time a certain amount of energy. This energy 
is transmitted very quickly to the plate through the edge of the 
chisel. The shorter and more rigid the chisel, the more quickly 
is the energy sent through the cutting edge into a portion of the 
plate. If it is not conveyed away rapidly from the edge, the 
amount contained in a small portion of material just imder the 
edge is very great, and the material is fractured there. As the 
energy of strain is proportional to the product of stress and strain 
or to the square of either stress or strain, the possibility of fracture 
for a material is represented by the square root of the strain energy 
it contains per cubic inch. If a material is brittle, there is a sort of 
instability which causes fi-acture at one place to extend to all neig'ii 
bouring places. And hence, if we deliver with great rapidity to a 
small portion of such a material a moderate supply of energy, it is 
sufficient to produce a large fracture. As our material becomes 
less and less brittle, we must have, over a larger and larger part of 
the volume in which we want fracture to occiu-, a sufficient supply 
of strain energy delivered. Hence, in cutting wood, we use a 
wooden mallet and a more or less lengthened wooden-headed chisel. 
The mallet and chisel act as a reservoir for the energy of the blow 
which is delivered to the wood from the edge of the chisel with 



APPLIED MSdHAKICS. 309 

comparative slowness and just in sufficient quantity to cause 
rupture in front of the edge. If the wood, without gaining in 
strength, became more rigid so as to be able to carry off more 
rapidly the energy given to it by the chisel's edge, it would be 
necessary to make the supply more rapid by using a more rigid 
chisel and mallet, and as we do this we must take care that the 
chisel itself near the edge is strong enough to resist fracture. This 
is one way of considering the effect of a blow. The exact mathe- 
matical consideration of what occurs in the impact of elastic 
bodies is not easy even for spheres and cylinders and other bodies 
of simple shape (see Art. 404). 

262. As we have just seen, the extra stresses due to loads 
suddenly applied are easy enough to understand. It is not 
so easy to comprehend why quietly varying loads which pro- 
duce no visible ^dbrations should produce what we call fatigue. 
The ordinary kinds of test as to strength under statical load, 
ductility, or elongation before fracture, applied to old rails, 
tyres, and other well-used material, show no great difference 
from what we obtain with new non-ductile material. Sometimes 
flaws are found, and it may be that fatigue somehow acts in pro- 
ducing minute flaws. The nature of the fractm-e in "Wohler's 
experiments is the same as that observed in old tyres and axles ; 
it shows no signs of ductility, and is as if the material were 
brittle. 



263. A piston-rod is subjected to tensile and compressive 
stresses, often repeated. It is found that its breaking 
strength is not 45,000 lbs. per square inch, which, let us say, 
it would be for a steady pull or push, but 15,000 lbs. per 
square inch. If, instead of such an action, we have a tensile 
stress which varies frequently from 30,000 lbs. per square inch 
to zero, the rod will break after a time. In the same way, 
steel which will bear a steady stress of 84,600 lbs. per square 
inch will only bear 46,500 lbs. per square inch if the stress 
varies between this and zero, but is always of the same kind ; 
whereas it will only bear 25,000 lbs. per square inch if the 
stress is sometimes a pull of this amount and is sometimes a 
push of the same amount. 



The above statement, the outcome of Wohler's experiments 
begun twenty-five years ago, and Fahbairn's experiments made very 
much earlier, was made fifteen years ago in the first edition of this 
book. The experiments made since give results of much the same 
kind. In the following table we have the most important results 
arrived at up to the present time, being the stresses in tons which 
require from five to ten millions or an indefinitely large number 
of apphcations of the load to cause fractm-e : — 



310 



APfLlED MfiCttAi^lCS. 
TABLE V. 





_2r-l J2 


Similar Stresses. 


One Stress 
Zero. 


Opposite 


Stresses. 


» 


Material. 


III 


















Least. 


Greatest. 


Least. 


Greatest. 


Least. 


Greatest. 




Wrought Iron. 


22-8 


12-0 


20-5 





15-3 


- 8- 


+ 8-6 




Krupp's Axle 


















Steel 


52-0 


17-5 


37-8 





26-5 


-14-1 


+ 14-1 


Ui 


Untempered 
















^ 


Spring Steel . 


57-5 


12-5 


34-8 





25-5 


-13-4 


+ 13-4 


Iron Plate . . . 


22-8 


11-4 


19-2 





13-1 


- 7-2 


+ 7-2 




Bar Iron 


26-6 


13-3 


22-0 





14-4 


- 7-9 


+ 7-9 




Bar Iron 


26-4 


13-2 


21-9 





15-8 


- 8-7 


+ S-7 


^ 


Bessemer Mild 
















CD 


Steel Plate ... 


28-6 


14-3 


23-8 





15-7 


- 8-6 


+ 8-6 


.H 


Steel Axle ... 


40-0 


20-0 


32-1 





19-7 


-10-5 


J.- 10-5 


Steel Eail ... 


39-0 


19-5 


30-9 





18 4 


- 9-7 


+ 9-7 




Mild Steel 
















p; 


Boiler Plate . 


26-6 


13-3 


22-6 





15-8 


- 8-7 


+ 8-7 





This exceedingly great weakening in material due to fatigue 
seems almost as if it had been vaguely known to English engineers 
from the beginning, and justifies the larger factors of safety which 
were wisely used in this country fifty years ago in railway bridges. 
Professor James Thomson showed that the two elastic limits ought 
to be called inferior and superior, as they are not necessarily equal 
positive and negative, but might even both be on the same side of 
the zero if the overstraining were great enough, and that variable 
displacements outside these limits would produce a destructive 
succession of sets. This theoretical deduction from the considera- 
tion of his overtwisted shaft has been completely verified by the 
experiments of Bauschinger, and may be said to completely explain 
the phenomena of fatigue. Bauschinger found that the elastic 
ranffe does not alter much, although either end of it may be 
altered, even, indeed, nearly to the ordinarj^ breaking stress. This 
change of either limit also is not very permanent, altering with 
hammering and other -s-iolent treatment. 



264. We can only refer to a few of the ways in which our 
principles are applied. When bolts tightly screwed up fasten 
two things — the flanges of a cylinder cover, for example — the 
bolts get lengthened and other things (let us call them the 
cushion) get shorten^-d. If extra forces are now applied it may 



APPLIED MECttAl^lCS. 311 

be that for a comparatively small extra lengthening of the bolts 
the cushion is nearly relieved and does not add its force as it 
did previously ; whereas in other cases, v^hen the cushion is 
more springy, there may be almost no relief, and the old forces 
due to the cushion may act along with the new forces. Yery 
often this initial tightening up of bolts is too great, and we 
have rule of thumb methods of designing sizes based upon an 
experience of the carelessness of workmen which it is difficult 
to express algebraically. One rule based on experience is that 
the length of a spanner shall never much exceed fifteen times 
the diameter of the bolt. Another, that, in certain kinds of 
machinery, bolts of less than a certain size shall not be used. 
Students ought to make careful sketches of the various kinds 
of bolts and nuts, including the usual forms of lock nuts and 
ways of locking, and also other kinds of fastening. In study- 
ing some of the proportions of these important details of 
machinery our theories are useful in suggestion; in some 
hands the theory is only a snare. Every true engineer must 
respect the proportions which have been arrived at by the fit 
and try, the failure and success methods of generations of 
engineers. When a very novel thing has to be made the good 
engineer sighs for practical guidance, and he is very cautious 
in using his theory. 

265. It is found then tliat wlien a rod is pulled, with, however 
small a force, not only does it get longer, but its diameter gets 
less. When, for example, a rod of glass is pulled so that its length 
increases hy the one-thousandth of itself, it is found that its 
diameter gets less by the one three-thousandth of itself. When a 
strut shortens, it also swells laterally. The ratio of the lateral to 
the axial strains in compression or in tension is called Poisson's 
ratio. It is of great importance in the theory of structures. 

The nature of the strain in a wire which is being extended, or 
in a column which is being compressed, cannot be said to be 
simple. If all lines in one direction, and in one direction only, 
became shorter or longer, the strain would be called simple, but it 
needs rather a complicated system of external compression or 
extension to produce this effect. No matter how a body is 
strained, if we consider a small portion of it we shall find that, 
besides angular changes, any strain simply consists of extensions 
and compressions in different directions. In fact, imagine a very 
small spherical portion of the body before it is strained. The 
effect of strain is to convert the little sphere into a figure called an 
ellipsoid (that is, a figure every section of which is an ellipse) or a 
circle. Eemember that every section of a sphere is a circle. It 
may be proved that there were three diameters of the sphere at 
right angles to one another, which remain at right angles to one 



112 



APPLIED MECHANICS. 




another in the ellipsoid, and are known as the principal axes of the 

ellipsoid. These directions are now called the principal axes of the 

strain existing at that part of the strained body. Along one of 

these directions the contraction or extension is less, and in another 

greater, than in any other direction whatever. 

Example.— Thx\B if m'n' (Fig. 173) is part of a long "wire 

subjected to a pull, the portion of matter which was enclosed in 

the A''ery small imagiriary 

spherical surface a b c d before 

the pull was applied is now 

enclosed in the ellipsoidal 

spherical surface a! c' ij' d'. 

The sphere has become an 

ellipsoid of revolution; A B 

becomes a' b', c d becomes c' d'. 

The strain in the direction 

. a' b' — A B -, ^, . . 

a B IS and this is 

AB , 

equal to the pull in the wire 
Fig. 173. per square inch divided by 

Young's modulus of elasticity, 
E. As, however, it is often more convenient to use a multiplier 
than a divisor, we are in the habit of using the reciprocal of e, and 
denoting it by the letter a. Thus, if the pull per square inch is p 
pounds, it produces a strain of the amount pa in the dii-ection 

A B ; the lateral contraction of the material is , and 

CD ' 

in this case is usually denoted by pfi^ the ratio fija being Poisson's 

ratio. 

266. In the following exercises on struts, the load is sup- 
posed to be carefully applied so that there is the same stress 
at every point of the section : — 

1. Find the area of the base of a sandstone column carrying a dead 
load of 6 tons. Take the ultimate crushing- stress as 3,600 lbs. per square 
inch, and use a factor of safety of 20. Ans., 75 square inches nearly. 

2. Find the least safe sectional area of a short cast-iron strut to bear a 
load of 12 tons. Take the safe stress as 15,000 lbs. per square inch. 

Ans., 1-8 square inch. 

3. The external and internal diameters of a short hollow cast-iron 
column are 10 inches and 8 inches respectively. If the safe working stress 
be 15,500 lbs. per square inch, find what load it will safely bear. 

If the external diameter had been 7 inches, what ought the thickness 
to have been to bear a load of 100 tons ? Ans., 196 tons ; -73 inch. 

4. A tie-rod made from |-inch wrought-iron plate has to sustain 
a load of 15 tons. What should be its width, allowing a working stress of 
7,000 lbs. per square inch ? Ans., 6*4 inches. 

5. The steam pressure in a locomotive boiler is 175 lbs. per square 
inch ; the stay-bolts which connect the flat sides of the firebox -^^th the 
end plate of the boiler are jjlaced 4 inches from centre to centre, vertically 
and horizontally. AVhat is the tensile force in each stay-bolt, and what 



APPLIED MECHAXICS. 313 

must be the diameter of each, the metal not heing subjected to a greater 
stress than 5 tons per square inch of the section of the bolts ? 

Ans., 2,800 lbs. ; 0-54 inch. 

6. What are the working and breaking loads <. ! pianoforte wire -^\ inch 
diameter ? And if the wire hangs vertically 5 miles, what weight at the 
end will, together with the weight of the wire, produce the working stress 
at the top? "WTiat is now the average load and the average stress in the 
wire ? By how much will the wire lengthen when subjected to it ? 

Ans., 297 lbs. ; 1,040 lbs. ; 27 lbs. ; 162 lbs. ; 16,460 lbs. per sq. in. ; 20 yds. 

7. A strut 100 feet long is made up of four angle irons of wrought 
iron (3| + 2^) ^, which are prevented from bending. Find the ultimate 
proof and working loads. How much shortening occui'S under the 
working load? Ans., working load 91,000 lbs. ; 0-28 inch. 

8. A piston-rod of mild steel 4 inches diameter, 6 feet long, the piston 
30 inches diameter ; what maximum pressure of steam may be used (1) if 
the engine is double acting; (2) if the engine is single acting? (See 
Art, 263.) What lengthening and shortening occur under these 
pressures? . Ans., 53 lbs. ; 160 lbs. ; (1) 0-08 inch ; (2) 0-12 inch. 

9. Columns about 8 feet high of brickwork, sandstone, and gTanite, 14 
inches square ; what working loads will they carry ? If they are 6 feet 
apart, and carry a wall of brickwork 28 inches thick, to what heights 
may it be carried? The brick columns being replaced by cast-iron 
cylinders 6 inches diameter, what ought to be their thickness? How 
much do they shorten under the load ? 

^?Z5., 15| tons each; 30ft.; 0'l2inch. 

10. The tight side of a gearing- chain taking a pull p, let it act at 5 
inches from the centre of the wheel, and transmit 20 horse-power at 100 
revolutions per minute ; find p. Each of a pair of links takes this pull. 
If the thickness of a link is one-third of the diameter of the pin, and if 
its breadth is two and a half times the diameter of the pin, find the section 
of each link. It is a good exercise to design the chain. 

Ans., 5,044 lbs, 

11. A single (a little over 0*2 inch thick) leather belt will stand an 
average pull of 1,000 lbs. per inch of its width. The weakness of the 
average fastener reduces this to about 300 lbs. per inch of width, and it is 
usual to take 66 lbs. as a working load. The pull on the tight side of a 
belt is two and a half times that on the slack side. The pulley is 30 
inches diameter, 150 revolutions per minute ; find the breadth of a single 
belt to transmit 5 horse-power, (See also the exercises Art. 185). 

A}is., 3 '5 inches. 
12. Columns of different material, constrained to keep of the same 
length, of cross-sections A;^, Aq, A3, etc, of coefiicients of expansion a^, 02, 
03, etc., unstressed, at a particular temperature, are raised t degrees in 
temperature. What is the fractional elongation? And what are the 
stresses in them ? 

Ans., If unconstrained, their fractional elongations would be ^a^, 
foo, etc. If a? is the real elongation, each has a compressive strain 
ttti — X, ta.2 — X, etc. Their compressive stresses are Ei {ta.^ — x), 
E2 (^Oo — x), etc., if E^, E2, etc., are their Young's moduli, and the 
total push ia each is AjEi {tax - ^)> H'^i (^«2 — ^)-> ©tc. The 
sum of these pushes is o (some of them being negative ; that is, 
tensile forces), or a^Ej {ta.^ - x) -^ k^-^^ {ta^ — «) + etc. = o, or 



314 APPLIED MECHAXICS. 



AiEjOi 4- iLiE.,a 



A^ Z^ -f A.-; Zo -h etc. ^ ' 11/ 

etc.. can "be found -wlien x is kno"^m. 

TiLiis, lor example, suppc'se there are columns equal in 
section, two of tliem of brass, one of cast iron. Then Aj = 2 a^, 
oj^ 19 X 10—^, 0,=: 11 X 10—^ for centigrade degrees ; e^ = 
9-2 X 106, E..= 17x106, 

_ ^ 2 X 9 2 X 106 X 19 X 10-6 4- 17 x 106 x 11 x ip- g 

■^ ~ ■ ' 2 X 9-2 X 106 -I- 17 X 106 

= tx lolo X 10—6. The compressive stress in the brass is 9*2 x 10'' 
(?19xl0-6-i 1.5ox 10-6) = ^x 30-2 lbs. per square inch. Tho 
stress in the cast iron is 17 x 10^ [t 11 x 10-6- ^ 1.5-.5 x 10-6) = 
— ^x 76-5. 

Thus, if ^ is 30 centigrade degrees, there is a compressiTe stress 
of 900 lbs. per square inch in the brass, and a tensile stress of 
2,300 lbs. per square inch in the iron. 

13. A bar of wrought iron 25 feet long and 1-75 inch diameter is 
heated to 180' C. While in this condition it is made to connect Idv 
means of nuts screwed on the ends) the two side walls of a building 
which have fallen outwards from the perpendicular. If the walls do not 
yield against the tendency of the bar to contract, find the pull between 
them when the bar has cooled down to 80' C. Take the mean coefficient 
of linear expansion of wrought iron as -0000124 for 1° C. 

Am., 92,3.50 lbs. 

14. Two bars of copi)er with a bar of wrought iron between them. aU 
of the same section and length, have their ends rigidly connected together. 
If the bars are heated from 15' C. to 100' C, find the stresses in the bars, 
and their fractional elongation, the coefficients of expansion for copper 
and wrouffht iron being taken at -0000172 and -0000124 respectirelv. 

^ Am., copper, 2,940 ; iron, 6.000 lbs. ; 0-00125. 

15. A tie-rod 100 feet long and 2 square inches in sectional area 
carries a load of 32.000 lbs., by which it is stretched f inch. Find the 
stress, strain, and e. 

Am., 16,000 lbs. per square inch; -000625; 25,600,000 lbs. per square 
inch. 

16. A wooden tie 40 feet long, 12 inches broad, 7 inches thick, was 
tested with a pull of 130 tons, which stretched it 1-2S inches. Find the 
value of E for the timber. Ans., 1.300,000 lbs. per square inch. 

17. A vertical wrought-iron tie-rod 200 feet long has to lift a weight 
of 2 tons. Find the area of the section and the diameter if the greatest 
strain is -0005 and e = 30.000,000. Xeglect the weight of the rod. 

A-m.. 0-293 square inch ; 0-616 inch- 
IB. Steam at a pressure of 200 lbs. per square inch is suddenly 
admitted upon a piston 18 inches in diameter. If the piston-rod be 3 
inches in dLLameter and 7 feet long, what is the compression and strain 
energy in the rod at maximum compression? e = 30,000,000. Find, 
also, the maximum stress in the rod. 

Ans.. 04 inch: 171 foot-pounds ; 14.400 lbs. -per square inch. 
19. A ship is moored by two cables of 90 feet and 100 feet in length, 



APPLIED MECHANICS. 



315 



respectively. The first cable stretclies 2f inclies, and the second stretches 
3 inches under the pull of the ship. Find the strain of each cable. 

Ans., -0024; -0025. 

The •weight of a rope in pounds per foot is taken as aff^ where p 
is its girth in inches, and its breaking weight in pounds is bff^ 
where a and b have about the following values : — 





a 


d 


Tarred Hemp 

White Hemp 

Iron Wire 

Steel Wire 


•04 
•04 
•13 
•13 


900 
1,300 
4,000 
6,700 



20. What lengths of themselves will each of these kinds of rope carry ? 

A)is., 22,500, 32,500, 30,770, 51,540 ft. 

21. Compare the weights and strengths of iion- and steel- wire ropes 
with iron and steel wires of the same cii'cumference. Afts., -49, "4, -49, '65. 

The Admiralty rule for the proof -load p in tons of the ordinary 
close-link chain of welded iron is 12d^, where d is the diameter 
of the iron in inches ; this means nearly 8 tons per square inch in 
the iron. For studded chain- cables it is p z= 18^^^ which means 
lli| tons per square inch in the iron. The working load is from 
half to one-quarter of this, depending on circumstances. The 
weight of either chain is about lOd^ lbs. per foot. Hemp rope of 
girth ^ is taken as being of about the same strength as a chain if 
g = 10^ to 11^. 

22. Two close-link chains, each making an angle of 50 degrees with 
the vertical, are to support a working load of 10 tons ; what is the proper 
size of the iron ? Ans., OS inch diameter. 

23. A ship of 2,000 tons (take it that one-quarter as much mass of 
water moves as the ship moves) is mo^dng at 0^1 knots, and is brought 
to rest in three seconds, the law of the motion during stoppage being 
«; = 0-1 COS. 7a, where t is time and k a constant, and v is in knots. The 
pull comes directly on a studded chain. If the chain gets its proof load, 
what is the diameter of the iron ? 



1 knot 



When three seconds elapse, cos. ki is 0, or cos. 3/c = cos- -, 

t-second units, t; = 0-1689 cos. "^t, because 

6 

or 1-6H89 feet per second, and acceleration 



or A = -. In feet- second units, t 
o 



6,080 



60 X 60' 



is — •lesg X - sin. -^t, being numerically -1 



X -, or 0-5 305 feet 



per second per second where greatest. The mass is 2,500 x 2,240 -;- 
32-2, or 173,900 in engineer's units. Hence the greatest force is 
92,300 lbs. 

24. Find the work which may be stored up in a pound of hard spring 



316 



APPLIED MECHANICS. 



steel when stretclied to its elastic limit, taking the modulus of elasticity 
as 35 X 10^ lbs. per square inch; the elastic limit, 100,000 lbs. per square 
inch; and the weight of one cubic inch, "29 lb. Ans., 410*5 ft.-lbs. 

25. A cylindrical rod of copper | inch diameter and 4 feet long, and 
one of wrought iron |- inch diameter and 3 feet long, are to be stretched 
the same amount. Compare the forces necessary to do this, the values of 
E for copper and wrought iron being 17,000,000 lbs. per square inch and 
29,000,000 lbs, per square inch respectively. Compare also the amounts 
of work expended in each case. Ans., 0*28 : 1, 0-28 : 1. 

26. What would be the resilience of a steel tie-bar 1 inch in diameter 
and 4 feet long if the bar becomes permanently stretched under a load of 
10 tons, the modulus of elasticity being 32,000,000 lbs. per square inch? 

Ans., 479 inch-lbs. 

267. Exercise. — Find every number in the following table. 
Values of tensile (ft-/2-E) or compressive (fc"/'2E) resiliences. 
These numbers express the relative values of the following 
materials for the making of springs in which elongation, 
compression, or bending occurs. In bending, the smaller of 
the two values, or possibly an intermediate value, must be 
taken. The numbers ///2n, or the shear resiliences per cubic 
inch, express the relative values of the following materials for 
the making of springs, such as spiral springs, in which shearing 
is most important. The numbers in each case show the 
amount of energy (in inch-pounds) which it is possible to store 
in each cubic inch of the material in the most carefully con- 
structed springs. In Art, 518 we give a statement showing how 
much less these stores usually are in ordinary springs. The 
work actually done upon ductile materials before fracture is 
often 1,000 times as great as the resilience, and in hard steel 
it is 150 times, in cast iron being twenty times the resilience. 





/r/2E 


/cV2e 


/.72v 


Cast Iron 

Wrought Iron 

Mild Steel (Hardened) 

Best Hard Steel 

Copper (Rolled or Drawn) . . . 

Fir 

Oak 


3 

10 

83 

600 

0-5 

4 

6 


12 
10 
83 

6*5 


5 

19 
128 
810 
0-73 



268. The diminution in bulk of a substance when it is sub- 
jected to pressure imiform all round, as, for instance, when it is 
surrounded by water in an hydi-aulic press, or sunk in the sea, 



APPLIED MECHANICS. 



317 



has been experimented "apon. The lessening in the hulk per cubic 
inch is called the cubical strain of the substance. The pressure 
in pounds per square inch all over its surface represents the stress, 
and it is found that the strain is proportional to the stress. In 
fact, in any substance the stress is equal to the strain multiplied 
by a certain number, for which the letter K is usually employed, 
called the Modulus of Elasticity of bulk. 

If an increase of pressure Bp causes the volume v to become 
V + 5'V (5y is usually negative), the stress being Sp and the. strain 

— Sv/v, the elasticity k. is defined by 5j» = — k dv/v, or k =: 

— V . dp/dv. In solids it is found that, whether the change takes 
place quickly (at constant entropy, as it is called in thermodynamics) 

, or slowly (at constant temperature), there is no very great differ- 
ence; but in gases and liquids it is very important to specify 
under what circumstances the rate of relative change of p and v is 
measui-ed. The ratio is in air 1-41; water, 1-004; alcohol, 1-22; 
ether, 1-58; mercury, 1-38; flint-glass, r004; drawn brass, 1'028; 
iron, 1-019; copper, 1-043.* 

The table, page 657, of slow moduli of elasticity of bulk is in 
pounds per square inch. 

269. A cube 1 inch in each edge (Fig. 174), subjected to a uniform 
compressive force of 1 lb. per square inch on the opposite faces, 
ADEP and BCLG. Evidently the edges 
AB, CD, LE, and GF, become I — a inch 
in length, a being the reciprocal of Young's 
modulus used above. Also the edges ad, 
B c, G L, and p E get the length 1 + )8 inch. 
If now we give to the faces a b c d and 
E F G L of this cube compressive forces 1 lb. 
per square inch, it is the edges af, etc., 
which shorten, and the edges ab, etc., 
which lengthen. Again, give the com- 
pressive forces to the third pair of opposite 
faces, A B G F and c d e l, and we have the 
edges AD, etc., shortening and b g, etc., 
lengthening. If, now, all three sets of 

compressive forces act at the same time (that is, the cube gets on 
every face a pressure of 1 lb. per square inch) , as the compressions 
and extensions are exceedingly small, each edge shortens by the 
amount a and lengthens by the amount 2 )8. Hence the edge 
which used to be 1 inch is now 1 — a + 2 iS inch. The cubic 
contents used to be 1 cubic inch ; it is now 1 — 3 (a — 2 ;8) with 
great exactitude. Hence 3 (a - 2 /8) is the amount of cubical 
strain produced by 1 lb. per square inch. That is, the Modulm of 
Elasticity of bulk, 

3 (a -2^)' 
and if we know a and jS it may be calculated. 

* Lord Kelvin (article on elasticity) gives the above, and also the follow- 
ing numbers. The ratios of the quick to the slow Young's moduli are : 
zinc. 1-008; tin, I'OOSG ; silver, 1-00315; copper, 1'00325; lead, I-QOSIO; 
glass, 1-0006; iron, 1-0026; platinum, 1-0013. 




Fig. 174. 



318 APPLIED ilECHANICS. 

Even very porous bodies, such as cork, have some elasticity of 
hulk. Fluids and homogeneous solids, such as crystals, 'are 
prohahly perfectly elastic as to bulk even at enormous pressures. 
Manufactured metals are generally porous, and alter (not 
necessarily increasing) in density after they have been hammered 
or drawn. 

Experiments on metals with great negative stress in all 
directions are wanting. Liquids do not seem capable of resisting 
great negative pressures, and the contrast between them and solids 
in this respect is remarkable. 

270. Students now know enough to make calculations on the stiff- 
ness and strength of ties and struts [only when the struts are kept 
from bending]. Before going on to other structures, even such 
simple structures as boilers and pipes, I wish to make a few 
remarks on the application of theory. 

The engineer must have some sort of theory to work upon. I 
shall give the theory recognised by men like Bankine, and also by 
the most successful practical engineers. It is simple as I shall 
give it. and fits fairly well a great number of practical conditions. 
It is foimded on the assumption that materials are homogeneous 
and not loaded beyond elastic limits, and yet it gives us knowledge 
which the judicioiis man finds useful- beyond elastic limits. It is 
also founded on the assumption that certain things, too difficult to 
calculate, are negligible, and hence its mathematical results ought 
to be tested by esperiment when this is possible. 

To give an example. Our theory of bending is founded on the 
assumption that the plane cross-section of a beam remains plane 
after bending. Every mathematical result seems to agree with 
every experiment made on beams, seeming discrepancies being 
always explainable by the tests not having been confined between 
the elastic limits of the materials. Again, the more elaborate 
theory of St. Tenant (Art. 311), invoh-ing fewer hypotheses, gives 
results that are practically in agreement with us. Henc€ we 
regard this easy theory which I shall give as one which may be 
depended upon in long beams, in spite of the fact that a plane 
section of a beam does not remain exactly plane. When it is 
applied in new cases not yet tested by St. Tenant's theory or by 
experiment, we use it only as a fairly trustworthy guide in our 
practical work. 

Another example. In the great twisting, which might occur 
without fracture ia indiambber shafts, it was known that the 
plane section of a square shaft did not remain plane. Xevertheless, 
the warping in ordinary shafts being small, a simple theory was 
adopted, in which there was the assumption of no warping. 
Results of experiments on round shafts agreed with the simple 
theory. Results from other shafts did not agree, and St. Tenant 
has shown us why there is a discrepancy. Although his investiga- 
tion is difficult to follow mathematically, his results are easy enough 
• to comprehend. I find it necessary, therefore, to give not merely 
the simple theory of the engineer, but an account of St. Tenant's 
results, and also, for advanced students, a short account of St. 
Tenant's theory of beams and shafts. It is to be remembered that 



APPLIED MECHANICS. 



319 



our theory assumes perfect elasticity of the material. "We shall 
see that at the bottom edge of a key- way in a shaft we have a 
place where the stress becomes very great indeed. I have made 
experiments on such a shaft with a key- way, and I find that it is 
by no means such a great source of weakness as the theory supposes, 
and this leads us to consider how the results of the theory are 
modified by the fl.ow of the material, instead of its fracture, under 
great stress. 

271. In thin-shelled vessels, such as boilers and pipes, sub- 
jected to fluid pressure^; inside, we assume that the tensile stress 
/is the same throughout the thickness ; so that if a is the area 
of metal cut through at any plane section of the boiler, af'm the 
resistance of the metal to the bursting of the boiler at that 
section. Now the equal and opposite force due to the fluid is 
A ^ if A is the whole area of this plane section of the boiler. 
Hence xp = a/ and ]? = a //a ... (1) gives us the bursting or 
working pressure if f is the ultimate or working stress. To 
prove this : — 

In Fig. 175 let F B c D E be part of a boiler whose separation 
from the rest by a plane section at f e we are now studying. 
Arrow-heads are drawn showing the forces 
with which the fluid everywhere acts nor- 
mally on the shell. We want to know the 
resultant of these forces. Imagine a boiler 
made with the part f u c D E and a rigid flat 
plate F E closing it. If we neglect the 
weight X)f the fluid, all the pressure forces 
on the shell balance one another. This is 
Newton's law of motion. The mutual 
forces of parts of a system cannot affect the 
motion of the centre of inertia of the whole 
system. (See Art. 482.) If the above boiler 
were placed upon a truck with frictionless 
wheels, there will be no more tendency to 
move on a level road when there is great 
pressure inside than when there is little, 
to pressure on any one little portion of the surface balances 
the forces on all the rest of the surface. Hence it is that if 
we make a hole there is a want of balance and our truck 
will move. When we make the hole the pressure everywhere 
changes because of the motion of the fluid, and hence we can 
only calculate the unbalanced force by knowing the momentum 
of the fluid which leaves the vessel per second. In the closed 




Fig. 175. 

The force 



320 APPLIED MECHANICS. 

vessel of Fig. 175 we know that the resultant pressure on the 
flat surface f E is A^, therefore the equal and opposite force on 
FBCD E is also A p. As an example of this we saw in Art. 
122 that the resultant force axially on the ram of an hydraulic 
press is exactly the same whatever be the shape of the end of it. 

Example. — Spherical boiler of diameter D. Any plane sec- 
tion is a circle. If we use the above rule for any such section 
we find that less pressure will burst the boiler if the section is 

diametral. The area of such a section is A = - c?^, and the area 

of metal laid bare is a = irdt. Hence (1) becomes p z=.-n:dtj 

-r|c?2or4^//^. . . (2). 

272. In a long cylindric boiler or pipe it is easy to show 
that (neglecting the strength of the ends) the tendency to hurst 
laterally is twice as great as the tendency to hurst endwise. 

Endwise, the area of a circular cross-section is A = -r d^, and 
' 4 ' 

the metal laid bare is tt dt, so that (1) becomes p = 4y7/d, 
just as in a spherical boiler. But laterally, at a plane passing 
through the axis of the boiler, a = ^ c? if Hs the length, and 
a = lit if we neglect the metal at the ends; and hence (1) 
becomes p = '^fijd . . . (3). The bursting pressure endwise 
being twice as great as this, we always take (3) as the formula 
for the strength of a cylindric pipe or boiler. 

Readers will now understand why in cylindric boilers the 
longitudinal seams are always much stronger than the girth 
seams. When the boiler has riveted joints we must, of course, 
regard the material as weaker than if it could resist tensile 
stress everywhere like a continuous boiler-plate. The working 
/for copper ought not to be taken greater than 2,400 lbs. per 
square inch for steam pipes. In cast-iron pipes and in steam- 
engine cylinders it has to be remembered that the difficulty in 
getting castings which are of the same thickness everywhere, 
and the allowance that must be made for tendency to cross- 
breaking when the pipes are handled, as well as the great allow- 
ance that must be made in steam-engine cylinders for stiffness, 
the difficulty of casting, and boring out, cause such calculations 
as the above to be somewhat useless. Thus it will usually be 
found that, whereas a large cast-iron water-pipe is not much 
thicker than the above calculation would lead us to expect. 



APPLIED MECHANICS. 321 

because it is usually carefully moulded in loam, yet a thin cast- 
iron pipe is often of more than twice such a thickness on the 
average, and it is our rule never to attempt casting a 9-foot 
length of pipe of less than f inch thick. In these cases the 
maximum/ for cast iron is taken as 1,500 lbs. per square inch, 
whereas for large pipes we usually take 3^000. 

EXERCISES. 

1. What must "bo the thickness of the plates used in the construction 
of a boiler 10 feet in diameter working under a pressure of 120 lbs. per 
square inch, taking the eflSciency of the joints to be 70 per cent, and the 
safe stress at 10,000 lbs. per square inch? Ans., 1-03 inch. 

2. A copper pipe is 4 inches diameter and f inch thick. What is the 
working pressure? Take/ = 2,000 lbs. per square inch. 

Am., 375 lbs. per sq. in. 

3. A vertical cast-iron pipe is 4 inches in internal diameter. The 
pressure at a certain place is 50 lbs. per square inch. At this place, and 
at places 100, 200, 300, etc., feet lower in level, find the proper thickness 
of the metal if the working stress/ is taken as 3,000 lbs. per square inch. 

Ans., -033 inch; -062 inch; -091 inch; -149 inch, etc. 

4. The rule used for loam-moulded cast-hon water mains is t — ^ + 
hd -j- 13,000 where h is head of water in feet, d diameter of pipe in inches, 
t the thickness in inches. A pipe 3 feet diameter, 1 inch thick ; find h. 
Find the corresponding pressure in pounds per square inch. Find what 
value of / the working stress will cause the ordinary rule for thin 
cylinders to give the same answer. 

Ans., 316 ft., 136-7 lbs. per sq. in., 2,460 lbs. per sq. in. 

5. A wrought-iron pipe 2 feet diameter, | inch thick ; its working 
stress is 5 tons to the square inch, but strength of plate is diminished 30 
per cent, because of riveted joint. What is the working pressure ? 

Ans., 326 lbs. per sq. in. 

6. A cylindrical boiler 12 feet diameter is constructed of -|f inch steel 
plate. The test pressure applied is 245 lbs. per square inch. Find the 
stress produced in the plate, and hence deduce the stress in the plate 
between the rivet holes, the sectional area being there reduced to -77 of 
the solid. Ans., 19,500 and 25,300 lbs. per square inch. 

273. Storage of Energy in Fluids. — The volume of a cylin- 
dric vessel being v and the safe pressure being p, we may take 
vp as proportional to the energy which may be stored. If the 
diameter is d and thickness t and length I, the volume is 

V = -r clH. Assume what is known to be true (see Art. 272), 

that the safe pressure for a long cylindric vessel is (neglecting 

the strength of the ends) p = — r, where f is the safe stress 

which the material will stand. The weight of the metal, 
neglecting the ends, is w = irdtlw, if w is the weight of unit 



322 APPLIED MECHANICS. 

volume of the material. The surface of the vessel is s = rdl. 
In all cases we neglect the ends. 

Then the storage capacity of energy per unit weight of 

vessel is -jdH-^ -^ irdtlw, or ^. So we see that it i,\ 

independent of the diameter. 

In water-tube boilers, therefore, which must store energy in 
this way, and where it is of importance that there should be 
great surface, we must consider surface -^ v]:). This is ^jtf ov 
4-/pd. Hence the thinner the tubes are, and, if the pressure 
is hxed, the smaller they are, the more surface they have as 
compared with their storage capacity for energy. 

The same considerations cause us to use thin tubes for 
surface condensation and other purposes, and there is the 
further consideration that accidents are less likely to be serious. 

In cases where energy is stored in hot water and steam, 
the rate of waste of energy is proportional to the surface, and 
this requires just the opposite conditions. 

EXERCISES. 

1. Sixty tuhes of wrought iron 4 inches inside diameter, 10 feet long, 
^ inch thick. Find volume, weight, internal area in square inches, and 
working pressure if working /= 10,000 lbs. per square inch. Neglect 
the ends. How many tubes 22 inches diameter, 10 feet long, will have 
the same volume? And find the thickness suitable for the working 
pressure. Find also the area and weight. 

Ans., 52-36 cub. ft., 6,729 lbs. wt., l2-o4 sq. in., 1,250 lbs. persq. in., 
2, 1-375 inches, 380 sq. in., 6,525 lbs. 

2. Cylindric boiler of mild steel 5 feet 6 inches in diameter, at elastic 
limit ; jDressure 300 lbs. per square inch. What is the thickness ? Joints 
supposed to be of 60 per cent, of strength of unhurt plate/. Eeplace this 
boiler with tubes 5 inches diameter of same length. How many tubes 
are needed to make up the same volume ? What will be their thickness 
(no seams) ? Their weight ? Now replace with 3-inch tubes, finding 
thicloiess and weight. 

Ans., -47 inch; 174 ; -021 inch ; 193 lbs. per foot length ; -013 inch ; 
199 lbs. per foot. 

274. When a belt or rope of weight iv lbs. per inch of its length 
is moving with a velocity of v feet per second in a curved path of 
radius r, the centrifugal force on a small length of it, r . Sd, is 

tv .)■ .59 —' Now, if a sketch be made showing how this force is 

Tff 

balanced by tensile forces t at the ends of the small length r . Sd, 
it will be seen that the centrifugal force is equal to t .Sd if 50 is 
very small ; so that t = wv'^lg, being independent of the radius. 
This, then, is the tensile force which acts in a belt or rope when in 
motion — an addition to the tensile force which acts when the ropo 



APPLIED MECHANICS. 323 

is at rest — and it must be taken into account in considering the 
strength, of a helt. Notice also that if a is the section of the rim 
of a pulley of wrought iron, the weight of it is •28« lb. per inch of 
its length. Hence the tensile force is •2%av^lg, or •28v-lff lb. per 
square inch is the tensile stress induced in the rim by centrifugal 
force when it moves at v feet per second. Taking the working- 
tensile stress in wrought iron of a pulley as 6,000 lbs. per square 
inch, the rim speed of a wrought-iron pulley ought not to exceed 850 
feet per second. The usual limiting speed of cast-iron pulley rims 
is 80 feet per second. Arms, if numerous, serve to diminish this 
action. If an arm of uniform cross-section moves at n turns per 

second, the limiting length l of it is — a / — — f,iiwis the 

2irn 'V ^ 
weight per cubic inch. Thus, if we take /ir: 6,000, w z= -28, 

916 
L = — . Thus, if n = 60 revolutions per second, l is about 

18 inches. If such arm has a section a at the distance r from the 
centre, it is easy to show that if each, section has simply to with- 
stand a pull due to the centrifugal force of the part outside it 
a = a^^-'br^, where, if r is in inches, b = Tr^wnh'^/ldSf. 

The condition as to strain of a rotating disc has been investi- 
gated by Dr. Chree (see Art. 307). 

As the tensile force in a perfectly flexible rope due to its motion 
is independent of the shape of the path in which a particle is 
moving, a very rapidly-mo^dng rope, if no external force such 
as those due to gravity act upon it, has no tendency to any 
alteration of shape ; each particle follows the path of every 
other, like particles of water in a stream-line in a state of steady 
motion. We can impulsively alter the shape of such a rope at any 
place, and then the new shape will be retained. Vibrations will 
be transmitted by such a rojDC as if along a naturally straight rope 
not moving in the direction of its length in which there is the same 
tension as is here produced by motion. A thinking student will 
from these facts readily see that there is a quasi-rigidity produced 
in the rope by the rapid motion. 

EXEECISES. 

1. Two puUeys, 3 feet 6 inches ia diameter, running at 150 revolu- 
tions per minute, are connected by a leather belt weighing 0*6 lb. 
per foot in length. Taking fx z= "3, find the greatest tension in belt 
when transmitting 7| horse-power. Ans., 260 lbs. 

2. In a travelling crane the driving rope runs at 5,000 feet j^er 
minute. Find the tension due to centrifugal action, having given that a 
rope 1 inch diameter weighs 0-28 lb. per foot of length. Ans., 60-4 lbs. 

275. Strength of Thick Cylinders. — Let the inside and outside 
radii be 1\ and r^, and the inside and outside fluid pressures be 
^jandj»o- Consider an elementary ring of metal, 1 inch parallel 
to the axis, inside radius r, outside radius r -f dr. Imagine a 
compressive stress p inside it and p -}- 8p outside {5p is usually 
negative in our examples), and a compressive stress q in t^e 



324 APPLIED MECHANICS. 

material {q is usually negative or the stress is tensile) at right 

angles to the radius, p x 2/- is a force tending to fracture this 

ring at a diametrical plane ; 2 {p + dp) (;• -f hr) — 2q . 5r is the 

force tending to prerent fracture. Note that there is a possible 

shear stress on the sides of this strip that we are neglecting, and a 

careful student -will give thought to the matter. Our justification for 

neglecting it lies in this, that the strength of our cylinder cannot he 

imagined to depend upon its length ; and if we consider a very long 

cylindric strip, end effects are negligible. Balancing the forces, we 

have the well-known rule for the strength of a thin cylinder. Divide 

by the thickness Sr, and imagine hr smaller and smaller, and we 

dp 
find p ■\- r — ^^ q . . . . {\). As the material is subjected to 

crushing stresses p and q in two dii'ections at right angles to one 
another in a cross-sectional plane, the dimensions parallel to the 
axis of the cylinder elongate by an amount which is proportional 
top + q. We must imagine this to remain constant if a plane 
cross- section is to remain a plane, and we make this reasonable 
assumption. Hence (1) has to be combined with 7; + ? = 2a... (2), 
where 2a is a constant. Substituting q from (2), in (1) we get 

-^ = — — — ....(3), and we find by trial that the solution 

is ^ = « -f — and q=- a - -^ . . . . {^), where the constants 

a and b are to be found by the conditions of any problem. Thus, in 

the case of a gun or hydraulic press, let the pressm-e inside be p-^ 

where r = ■?•-, and let P(^^^ where r = r^^. If we insert these 

conditions in (i), we find {q is everywhere negative, and I shall 

use /, the tensile stress, to replace — q) 

_ f = r,2 r,^ + r-2 

^ J ^1 ^^2 _ ,.^2 ,.2 .^}- 

r '^ I r ^ 
/ is greatest at r = 9\, and is then /^ = p^ "^ ^_^ .... (6). 

The student is to note that the circular compressive strain at 
any place is qa -j^fi- This is the fractional diminution of the radius r. 

A student ought to take an example such as : An hydraiilic 
press has an inside radius i\ =: 4 inches ; the stress is not to 
exceed 5,000 lbs. per square inch; find the gTeatest possible 
pressiu^e pi^ first, if the thickness is 1 inch, then if the thickness is 
2 inches, and so on. ISTote how little gain there is by increasing 
the thickness more than a certain amount ; and it may be well to 
write out a list of numbers for various thicknesses, showing among 
other things the gain in weight. 

EXEECISES. 

1. In a certain kind of work either one cylindric hydi-aulic 
press of 24 inches diameter or four are needed, of same aggregate area 
and same material, to stand the same pressure. Compare the square area 
on which the two arrangements will stand. Observe that the ratios of 
nternal to external radii will be the same in the small and large presses. 



APPLIED MECHANICS. 325 

If llie student figaues it out, he will find that the four will just occupy the 
same square space as the single press, and they will weigh just the same. 

2. Thin Cylinder. — Take 1\ = ^d and Tq = \d + t, where d is the 
inside diameter of a cylindric holier and t is its thickness, and assume that 
t is small. Then (6) hecomes 

f=P ild^ + ^^ + i')l{dt + t^) =pd ( 1 + ^ + 1^^ I2t (l + . 

A first approximation which is generally used, and has been given above, 

is/ = |^ . . . . (7). A second is/=ff + ii^ • . . . (8). 

3. A gun of 12 inches internal and 24 inches external diameter is 
subjected to a maximum internal pressure of 40,000 lbs. per sqiiare inch. 
Find the stress produced at r = 6, 7 J, 9, 10^, and 12 inches. Find what 
was the initial stress everywhere if it was just sufficient to caiise the final 
stress everywhere to be the mean of the stress produced at r "^ 6 and 
r = 12. ISTow make diagrams showing the state of stress when Px = 
60,000, 50,000, 40,000, 30,000, 20,000, and 10,000 lbs. per square inch. 

Ans., 66,666 ; 47,466 ; 37,037 ; 30,748 ; 4,444 lbs. per square inch. 

4. Pipes of a water-pressm^e supply company are to withstand a 
possible pressm-e of 1,000 lbs. per square inch; they are of 6 inches 
internal diameter. What is the outside diameter, the safe tensile stress of 
the metal being 3,000 lbs. per square inch? Ans., 8-485 inches. 

5. Pipes are to withstand a working pressure of 1,000 lbs. per square 
inch. If their internal diameters d are 2, 3, 4, 5, and 6 inches in each 
case, find the thickness. Find in each case the weight w per foot length. 
Draw a curve shawing the relative values of iv and d. 

Ans., thicknesses 0-41, 0-62, 083, 104 inch. 

6. A cast-iron water-main is 30 inches internal diameter and 1 inch 
thick. What is the greatest head of water that it ought to be subjected to ? 
Safe tensile stress, 3,000 lbs. per square inch. Numbers to recollect are : 
34 feet of head represent 1 atmosphere, or head in feet — 2*3 = pressure 
in lbs. per square inch. If the pipes are wi'ought iron, what ought their 
thickness to be if safe /= 10,000 lbs. jDcr square inch, and if the longi- 
tudinal seams are of 60 per cent, of the strength of the unhurt plate ? 

Ans., 460 ft.; -5 inch. 

276. In the above theory we have considered the material 
initially unstrained ; or, rather, the stresses and strains calculated 
by us are additional to any initial stresses and strains in the 
material. 

The student wiU see why the outer material of a thick cylinder 
is comparatively useless if he shows in a cur\e/for various values 
of r, calculating from (5), for / decreases as the inverse square of r. 
If, when Pf^ = Pi = 0, there are already strains and stresses in the 
material, the stresses given in (5) are algebraically added to those 
already existing at any place. Hence, in casting an hydraulic 
press, we chill it internally, cold water ciixulating in a metal core 
painted with loam ; and in making a gun we build it of tubes, each 
of which squeezes those inside it. So that there is considerable 
compressive stress where r = 7\ and considerable tensile stress 
where r = Tq before any pressure comes on inside. We try to 



326 APPLIED MECHANICS. 

create such initial stresses that when there is the maximum 
pressure p-^ the material has about the same tensile stress in it 
everywhere. Much knowledge is needed to produce this result in 
guns. 

Exercise. — Thick spherical shell suhjected to iaternal fluid 
pressure. If p is the radial compressive stress at a point at the 
distance r from the centre, and q is the tangential tensile stress 
there, show after the manner of Art. 275 that jt? = a -f 2B/r3, 
q = A — B/r^, where a and b are constants, which may he 
found if the internal pressure and the inner and outer radii are 
stated. 

277. The above theory of the strength of thick cylinders seems 
to agree with our practical experience for such ratios of r^ and r^ as 
we find in the pipes and presses used by engineers. But all the 
rules given above show that a flat plate has no strength. The 
neglected terms in our theory become important in this case. In 
truth, the mathematical theory of a shell is so troublesome that we 
cannot say there is yet a satisfactory treatment of it. The strength 
of a flat plate has, however, been investigated in a niunber of 
cases, and we are led to the following results : —For a circular 
plate of thickness t and diameter d supported all round its edge, 
with a normal load of p lbs. per square inch, if / is the greatest 
stress in the material, / = br'^plQ>P. If the plate i^fxed all round 
its edge, /== Ir^plZP. A square plate of side 5, fixed at the edges, 
/= s^plA:t . A rectangular plate of length I and breadth b, fixed 
round the edges, /= I^b'^p/2i"{l'^ + ^^). A round plate with a 
load w in the middle, supported at the edges, f= ^\|■^■t'^. For 
stays in square formation, distance asunder being s ; each stay 
has a load^s^^ ^nd the greatest stress in the plate of thickness t 
is Is^pi'qfi. 

278. When the pressure is greatest outside a thin shell, its 
strength to resist collapse ought evidently to follow the law (1), 
which becomes (3) if the vessel is cylindric ; but it is in the very way 
in which a strut may be relied upon if the slight lateral restraints 
are provided which prevent bending. So also the thiu shell of a 
boiler flue must be provided with certain restraints against 
buckling ; and just as we have (Art. 372) a theory of laterally 
unsupported struts, so we have a theory of long boiler flues. 
Beyond a certain length for a given pressure there is instability, 
and hence flues are either corrugated or furnished with a number 
of rings. The most important practical outcome of the theory is 
that the distance between two rings divided by Vdi must not 
exceed a certain limit. The rules usually followed by boiler- 
makers are given in the new edition of my book on Steam. 

Exercise. — In a corrugated flue of 3 feet mean diameter the 
plate is -i- inch thick, but the corrugations, being longer than the 
axial length, make it virtually | inch thick. What is the working 
pressure if the working compressive stress (we allow for corrosion, 
etc) in the material is 3,000 lbs. per square inch ? 

Ans. , 125 lbs. per square inch. 



APPLIED MECHANICS. 



327 



MORE DIFFICULT EXERCISES ON THICK CYLLNDERS. 

1. A tube of wrought iron, inside radius 3 inches, outside 4 inches, 
outside pressure 0, What is the inside pressure p to produce a maximum 
tensile stress of 15,000 lbs. per square inch ? Find the fractional increase 
in size of the inside radius. Here ]} = where r = i; p = f, and q ~ 
- 15,000 where r = 3. Inserting these values in (4) Art. 275, we find 







15,000 = b( 






h = 86,400 


q = a- hfr^ 


b 

' 9 




- 15,000 = a - 


a= - 5,400 



G + re) = 



25_ 
144 



p = - 5,400 + 9,600 = 4,200 lbs. per square inch. 
Let the student calculate — q for several values of r from 3 to 4, and 
plot his results on squared paper. 



r 3 

i 


3-25 


3-5 


3-75 


4 


-q 15,000 


13,580 


12,453 


11,544 


10,800 



The fractional diminution in size of any radius is qa - p 13, or 



(a - iS) fl^ - -2 (« + ^)- Taking- a = ^ x 10 



g .. _„ 7^^^ ^10-7^thefi'ac- 



tional diminution of the 3 inch radius is 



10 



7 * 5 ,^, 



^ = - 5,350 X 10-7. 



That is, the 3 inch radius increases, becoming 3 x 5,350 x 10-'', or 0'00161 
inch larger. 

2. A tube of wrought iron, inside radius 2 inches, outside 3 inches, no 
pressure inside; pressure _^ = 4,200 lbs. per square inch outside. Find the 
cii'cular compressive stress everywhere, and also the diminution of the 
outer radius. 

Herein i^ = « + ^ » ? = « - ^ • • - • (!)• 

Fractional diminution \ ^ / , o\ /o\ 

of the 3 inch radius ) = (« - ^) « - 9 (« + p) (^) 

\\e have p = 4,200 where r = 2,; p = o where r = 2. 



4,200 = a -\- 



= a + 



4,200 = s(l-i)=- 
b= - 30,240, a = 7.560 



328 APPLIED MECHANICS. 

Hence q = 7,560 + Z0,2i0Jr^. Thus for the following values of r we have 



r 


2 


2-25 


2-5 


2-75 


3-0 


a 


15,120 


13,533 


12,400 


11,560 


10,920 



And the 3 -inch radius decreases hy the fraction of itself 



7,560 X 10' 



' + ^%^- 



7, 



or 3-29 X lO""*, so that the 3 inch radius becomes -00099 incli 
smaller. 

3. A tube of radius 4 inches outside, and radius 2-9984 inside, is 
squeezed in some way upon a tube 2 inches inside radius and 3-00098 
inches outside radius. Find the compressive circular stress at all points 
in both tubes. 

Ans., It will be seen that I have taken just the sizes necessary to 
produce the states of Exercises 1 and 2. It is evident that we may take 
the outside radius of the inner tube as 3 inches, and the inside radius 
of the outer tube 2 '9974 inches. Observe that as the coe£&cient of ex- 
pansion of iron is 1*2 x 10 ~, to produce a fractional increase in size 
•0026 -j- 3, the outer tube must be raised in temperature more than 

— X — -i- (1 2 X 10 ~ ^), or 72 Centigrade degrees, before it wiU slip over 
o 

the other. 

4. A tube of wrought iron, 2 inches radius inside, 4 inches outside, is 
subjected to a fluid pressure of 50,000 lb. per square inch inside and no 
pressure outside. Find the tensile stress everywhere in the material or 
the values 0,1 -q. 

Here in p=^a-\--^^q — a — -^, insert p = 50,000 where r = 2, 

Ci p f* C C*7 

p = where r = 4, and so find - q = '-^ h 16,667. Of course 



- ^ is the tensile stress — 








r 


2 


2^ 


3 


3^ 


4 


-Q 


83,333 


59,333 


46,300 


38,440 


33,333 



5. To sum up our results. The built-up tube of Exercise 3, with initial 
tensile stress/ or -qof E-t. 3 taken from the tables of Exercises 1 and 2, is 
subjected to the internal fluid pressure of 50,000 lbs. per square inch of 
Exercise 4, there being no pressure outside. We have seen that/' ox -q 
of Ex. 4 shows the tensile stress produced by the fluid pressure, and 
hence/", which is /' -}-/, is the real tensile stress every where in the 
compound tube. Students ought to draw curves showing/,/', and/". 



APPLIED MECHANICS. 



329 



r 


2 


2k 


2* 


21 


3 


3 


3i 


3J i 3^ 


4 


f 


-15,120 


- 3,533 


- 12,400 


- 11,560 


- 10,920 


15,000 


13,580 


1 
12,453 ' 11,544 


10,800 


f 


83,333 


69,337 


59,333 


51,937 


46,300 


46,300 


41,317 


38,440 35,727 

1 


33,333 


f" 


68,213 


55,80i 


46,933 


40,377 


35,380 


61,300 


54,897 


50,893 47,271 


44,138 



279. These exercises will enable the student to understand the 
usual calculations of shrinkage which, must be made in building up 
a gun. He will have no great difficulty in working out all the 
necessary formulae himself, but a man interested in gun-making 
ought to refer to three articles published in Nature about August, 
1890, by Prof. Greenhill. It will be noticed that when a gun is 
built of tubes there is a sudden change in the tensile stress at the 
common surface of two- tubes. To get a more uniform stress 
throughout, instead of using many thin tubes we now use an inner 
tube of steel strong enough to withstand the longitudinal forces, 
then a thick layer of steel wire wound on with varying tension, 
and then a covering tube, which is almost unstrained initially. 
Probably the hydraulic presses of the future will also be built up 
in this way. 

Prof. Greenhill saj^s : " Mr. Longridge's principle of strengthen- 
ing a tube with wire wound with appropriately varying tension 
will be found useful in peace and in war. He can claim credit 
that a gim strengthened on this principle (the 9-2 inch wire gun) 
was chosen, from its great strength, to test the extreme range of 
modern artillery in 1888, with what were called the "Jubilee 
rounds," when, with an elevation of about 40 degrees, a range of 
21,000 yards, or 12 miles, was attained, the projectile weighing 
380 lbs., and the muzzle velocity being abou.t 2,360 feet per 
second." 

280. The following example will illustrate how calculations 
may be made : — 

Example. — A steel tube 4 inches internal and 5 inches external 
diameter has steel strip wound on it to the external diameter of 10 
inches, under a constant tensile winding stress w lbs. per square 
inch. Find the radial and circumferential stress anywhere. 

Consider the place r in the wire. When the wire has already 
been wound to an outside radius p, let a new layer of thickness Sp 
be wound on. This produces a pressure to . dpjp. It is easy to 
see, just as in Example 2, that a radial pressure tv . dp/p at p, there 
being no radial pressure at r^=r 2", produces at r an increase Sp 
of radial compressive stress, and an increase 5q of circumferential 
compressive stress where 



330 



APPLIED MECHANICS, 



W . dp/p 



(1)> 



(2). 



These effects are produced by winding on a new thickness So of strip. 
Integrating these from p = r to p = Tf^, we find 

where x stands for r^/n^ and Xq for r^lr-^. The tensile stress in 
the strip or wire was w lbs. per square inch, and we see that this 
is now reduced to w; - ^ =/ where 

/=.{l_i(lH-l)log.^}....(5). 

T have used (3) and (5) in calculating the following numbers : — 



r 


2-5 


3 


Zb 


4 


4-6 


5 


P 


•402 w; 


•399 w; 


■Zlbw 


•210w; 


•103w 





f 


- SSSw 


- -036^7 


•382 m; 


•65w 


'847 w 


w 



At r = 2J inches, the outside of the steel tube and inside of 
the wire, there is a discontinuity. Calculating- from (3) and 
(5), the tensile stress there is/= — 0*832 w; and 2^ = 0-402 w. 
Now the metal of the solid tube is subjected to external pressure 
•402 w and no internal pressure, and we find that in it the 

compressive circumferential stress q = 4:-4:'J w (i + -2) • • • • (6), 
and the radial stress is^ = 4-47 «t; ( * ^2 ) • • • • C^)* 



I have calculated the values of p and g, or rather of 
I call/, from (6) and (7) for the solid tube : — 



q, which 



r 


2 


2i 


^ 


-S=f 


- 2-235 w; 


— 1w 


- 1-83 w 


P 





•234: w 


•402 m; 



APPLIED MECHANICS. 



331 



We know now tlie internal condition of this coiled structure, 
which, we may call a gun. 

Let the gun, considered as homogeneous, Tq = 2, r^ = 5, 
he suhjected to fluid pressure 3 w inside and no pressure outside. 
It is easy to see that this fluid pressure would produce the tensile 

stress everywhere ('57 + — ^j— ) w, and as hefore, leaving out the 

w, the numbers marked /' in the following table show its value. 
In this table the initial radial compressive stress 2> and circular 
tensile stress / are copied from the previous tables. The actual 
tensile stress, therefore, in the gun is/' + /, shown as/". 



r 


2 


H 


24 


2i 


3 


3i 


4 


H 


5 


P 





•234 


•402 


•402 


•399 


•315 


•210 


•103 


C 


f 


-2-235 


- 2 


-1-83 


-0-833 


- ^036 


+ •382 


+ •65 


+ •847 


1 


f 


4-14 




2-85 


2^85 


2-16 


1-74 


r46 




•14 


f" 


1-90 




1-02 


2-02 


2-12 


2-12 


2^11 




2^14 



It will be observed in the above table that the actual tensile sti-ess in 
the wire is nearly constant. This is a happy accident. 

AVe usually assume that/" shall be a fixed constant in the Avire ; /' is 
known, so we may calculate the necessary values of /, the initial tensile- 
stress everywhere, in the finished gwci, /is evidently of the shape /=m- 
hjr-. To produce this particular /, it is necessary to alter the winding 
tensile stress w in the wire as the winding proceeds. Thus in (1) and (2), 
if, instead of w being constant, we have tv a function of p ; remembering 
that 10 is to be looked upon either as a function of r or p, integrating (2) 
and writing what is equivalent to (5), we have 



/ 1 1 \ r iv.pjp 



f 



Differentiating with regard to r, we see that if d be written for w -/ and 
if X stands for r^jr^, then 



a;''^ + 4^-l 



+ / 



x+l 



= 0. 



dx ' x{.v--l) •' 'lx{x-\) 
It is not difficult to solve this and so find the way in which %v, the 
♦vinding tension, ought to vary. 



332 



CHAPTER XIV. 



SHEAR AND TWIST. 



281. Let c d (Fig. 176) be the top of a firm table, f h a long 
piism of indiarubber glued to the table, a b a flat piece of wood 
glued along the upper side of the indiarubber. We try in this 
way to apply a horizontal force to the whole upper surface of 
the indiarubber, so that if, for instance, the pull in the cord is 




Kg. 176. 



7«_0i.-_<. 



20 lbs., and the upper surface of the indiarubber is 10 square 
inches in area, there will be a force of 2 lbs. per square inch 
acting at every part of the surface, and this force will be 
transmitted through the indiarubber to the table. When the 
length of the prism is great compared with z f, we may suppose 
that the bending in it is very small, and in this case we say that 
the indiarubber is being subjected to a simple shear strain ; the 
force per square inch acting on its surface is also acting from 
each horizontal layer to the next, and is called the shear stress. 
If YOU had drawn vertical lines like y'x before the cord was 
pulled, you would now find them sloping like Y X. Thus, 
making a magnified drawing of y x in Fig. 176, the point y' 
has gone to y, and any point like m has gone to N. Points 
touching the table cannot aiove, but the farther a point is away 
from this fixed part the farther it can move. Now suppose 
that y' y is O'Ol inch, and we know that x y' is 2 inches, what 
is the amount of motion of M if M x is 1 -7 inch 1 Evidently 



APPLIED MECHANICS. 333 

y' y is greater than m n just in the proportion of y' x to m x, or 
2 to 1*7 ; hence mn is 0-0085 inch. Thus the motion of any 
point is simply proportional to its distance above the fixed 
plane, and if we know the amount of motion at, say, a distance 
of 1 inch, we can calculate what it must be anywhere else. 
The amount of motion at one inch above the fixed jolctne is called 
the shear strain. The motion is small, and it is evident that 
the shear strain is the angle y' x y in radians. In this case we 
have supposed the force on f g to be 2 lbs. per square inch. 
This is said to be the amount of the shear stress, and it pro- 
duces or is produced by a shear strain whose amount is '005 
inch per inch. If the shear stress were 4 lbs. per square inch, 
you would find the strain to be -01; if the stress were 8 lbs. per 
square inch the strain would be -02. In fact, we find experi- 
mentally that the stress and strain are proportional to one 
another. Thus if, instead of indiarubber, we had a block of 
tempered steel, we should find that the force in pounds per 
square inch is equal to 13,000,000 times the strain. This 
number is called the modulus of rigidity for steel j it is given 
in Table XX. It seems a pity that the name shearing elasticity 
is not given to this number. 

282. However long we may make our block of indiarubber in. 
Fig. 176, we shall still have some bending in it ; that is, the stress 
will not be uniformly distributed over each horizontal layer (see 
Chapter XXI.). To prevent this bending- effect, and to produce a 
really simple shear strain, we ought to have force distributed over 
the ends f z and g h 
of the same amount 
per square inch as F 

we have now acting 
over p G and z h. , 

These are shown in / 

Fig. 177. Where p , / 
is the pull in the cord ^ ^ 



of Fi2-. 176, r' is the ~ i 



equal and opposite ' ^ 

force exerted by the p.^^ ^^^^ 

table on the glued °* 

underside of the india- 
rubber, and F and f' are equal and opposite forces distributed 
over the ends, such that the couple ff is able to balance 
the coux^le p p'. There can now he no bending moment at 
any place. As f multiplied by the length of the prism is the 
moment of the couple f f', and is equal to p multiplied by the 
vertical dimension, we see that p distributed over the horizontal 
surface is the same stress per square inch as r distributed over the 



334 APPLIED MECHANICS. 

ends. From such a material then, if we cut a cubical block a 
(Fig. 178), its horizontal faces Yy and x a? are acted upon by- 
equal and opposite tangential forces, and its faces y x y x are 
acted upon by forces of exactly the same amount. The faces 
parallel to the paper have no forces acting on 
lx>suv^ n^ them. This will give you the best idea of simple 
' \ shear stress. The material in Fig. 176 near the 
/ ends of the block does not get a simple shear ; 
; t but if the block is very long, then at the middle 
there is a nearly simple shear acting. 
— ^ In Fig. 178 the cube x y' y' x has become 

^^ TLxyx. Suppose the side of this cube to be 1 

Fig. 178. inch, then y' y is the shear strain, which I shall 

call s. The tangential force distributed over 
Y 2/ is i? lbs., let us say. Then, if we denote by the letter n the 
modulus of the rigidity of the material, p = n s. 

283. Examine. — A beam of steel has one end fixed, and at 
the other is a weight of 20 tons. The cross section of the beam 
is 2 square inches in area, and the length of tlie beam is 5 inches. 
Besides the deflection of this beam due to bending, there is a 
certain deflection due to shearing ; how much is it % Answer : 
the shear stress is 10 tons, or 22,400 lbs. per square inch. 
This produces a shear strain of 22,400 -f 13,000,000, or -00172. 
This is the amount of yielding at 1 inch from the fixed end, and 
at 5 inches the yielding must be 5 x '00172, or -0086 inch. 

In a short beam like this, or in one 20 inches long, if we con- 
sider, for example, that it is 1 inch broad and 2 inches deep, we may 
calculate the deflections due to bending and to shear, and reflect 
upon the fact that in very short beams the yield due to shear 
is much more important than the yield to bending ; whereas in 
long beams we may, and indeed always do, neglect altogether 
the deflection due to shear. These reflections are in the main 
correct, but the actual distribution of shear stress over the 
sections of a beam not short and not long is unknown to us. 
Its distribution in the section of a long beam will be given in 
Art. 369, and it is obvious that the calculation of the deflection 
due to shear is not so simple as in very short beams. 

284. The shear stress which will produce rupture is not 
well known for any substance except cast- and wrought-iron, but 
the shear stress which will produce permanent set is fairly 
well known, and we are also agreed as to the ordinary working 
shear stress of materials. For wrought iron and mild steel it is 
usually regarded as from 75 to 85 per cent, of the tensile stress ; 
but in a single-riveted lap joint in boiler-plates, as the holes are 
usually punched (and this weakens the metal), and as rivet iron 



APPLIED MECHANICS. 335 

is usually of a better quality than plate, the cross section of 
the iron which is left, which is resisting pull, is made to have the 
same area as the cross sections of all the rivets, which, of course, 
resist shearing. Besides breaking by either a tensile or a shear 
stress, a riveted joint may give way by the rivet crushing or 
being crushed by the side of its hole. Again, in many riveted 
joints, when the rivets are long, as they tend to contract in 
cooling and are prevented by the plates, so much tension may 
remain permanently in them that they are greatly weakened. 
In bolts there is usually some bending, and consequently a want 
of perfectly uniform distribution of the shear stress, and they 
are made larger than rivets in the same positions. 

285. In the punching of rivet-holes it may be taken that a 
shearing force v acts on the material ; the area of the curved 
side of the hole, multiplied by the breaking shear stress of the 
material per square inch, represents the force with which the 
punch must be pressed down on the plate. The punch must be 
able to resist this force as a compressive stress on its own 
material. Experiments made on punching-machines show that 
about 24 tons per square inch is the average shearing force 
require'd. This pressure has to be exerted through a very 
short distance. In shearing-machines, if the entire edges of 
the shears coincided with the plate as soon as they touched 
anywhere there would be the same soi't of effect produced ; 
but by inclining the edges the shearing action does not occur 
instantaneously at every place, and the rupture being more 
gradual than in punching, the shearing resistance is usually 
from 10 to 30 per cent. less. It is very probable that the 
power lost in punching- and shearing-machines is wasted rather 
in the friction of the heavy parts of the mechanism than in the 
&lmost instantaneous effort of cutting the material. TI1-3 effort 
required seems rather that of an impact than of the more 
gradual action to be found in most existing machines. At the 
same time we must remember that M. Tresca's experiments 
indicated a flowing of the metal. Machines such as hydraulic 
bears and shears may be uneconomical as to mere energy, but 
they produce the higher economy due to convenience and 
ceitainty. A man can stop the motion of the hydraulic punch 
very rapidly if he sees that the plate is not quite right in posi- 
tion. In the fly presses used for hand punching, and used largely 
in coining, the idea of an impact is already in use ; it will come 
much more into use in large machines when engineers become 



336 



APPLIED MECHANICS. 



better • acquainted with the distinction between forcfi and 
energy. 

286. Riveted Joints. — Students are supposed to have made 
sketches of a number of well-proportioned joints. In applying 
the results of experiment and our imperfect theory to actual 
structures we must remember that our practical conditions some- 
times closely agree with our hypothetical conditions, and then 
our calculations may be fairly exact ; whereas in other cases our 
theory is only an imperfect sort of guide in our workshop 
systems of seeking for success and avoiding failure. It is only 






I Fig. m I 

a very capable engineer who knows exactly what weight ought 
to be given to his theory in every case. 

If we had a perfect theory we need only consider the various 
ways in which a riveted joint may fracture. Then we should 
state the algebraic conditions that the joint shall be equally 
ready to fracture in these various ways, and we have at once 
the right proportions. 

Thus, take the strip which illustrates the single riveted lap 
joint of a boiler plate, Fig. 179 (1), or the single riveted butt 
joint of Fig. 179 (5). a b = j9, the pitch or distance from centre 
to centre of rivets ; t is thickness of plate ; X is d c^ the overlap 



APPLIED MECHANICS. 337 

minus radius of rivet ; f^ the shearing stress which the luvet 
will stand ; f^ the tensile stress %Yhich the plate will stand. 
If fracture occurs as in (1), the tensile force p which the 

joint will stand is p = r; d^J& .... (1). If in double shear, as 

in (5), P = \d^f, .... (5). If, as in (2), p = (^^ -d)tf,.,, . (2). 

If as in (3), when the iron of the plate is crushed by the rivet 
or the rivet is crushed by the iron, even the most absurd 
person will hesitate when he puts F=f^td .... (3),/c being 
some kind of surface crushing resistance per square inch, and 
td being assumed somehow to represent the surface at which 
crushing may take place. If, as in (4), we assume that the part 
E G F breaks like a beam fixed at the ends and loaded in the 
middle, we have a much wilder assumption, giving p = X^tft -r- 
^d . . . . (4). Now we cannot assume that p has any of the 
above values. There is always great friction at the joint ; the 
rivet is in a state of unknown tension, and we have no inform- 
ation as to y's under such circumstances, and in a lap joint 
there is evidently a bending action due to the plates not being 
in line. Nevertheless, all the above formulae and similar 
formulge easily made out for all joints may be made use of to 
guide us in obtaining information when we vary proportions 
and make tests. 

For example, tests may be made upon A. and d, and it has 
been found that \ from the edge of the hole to the edge of the 
plate must be at least equal to d, and considerations of economy 
and workmanship guide us in adding about a quarter of an inch 
when only half-inch rivets are used. Now if we put (4) equal 

to (3), or \Hf, ^ Id =f,fd (6), we have \ = d.^y^sfc^ 

Jx 
It is dangerous to follow this any further ; we have reached a 
rule (by taking convenient values for the stresses) that agrees 
so well with the practice of the best makers that we shall be 
apt to think our theory more valuable than it is. It will be 
noticed that if we write (3) equal to (1) we get a rule which 
also agrees fairly well with practice ; but if v/e write (4) equal tc 
(5), which we have just as good a right to do, we find that \ 
ought to be ^'2 times as great when the rivets are in double 
shear, and this is certainly not in agreement with good practice. 
If we imagine that instead of (4) we use the idea of a beam 



338 APPLIED MECHANICS 

breaking by shearing, we again find that X ought to be propor- 
tional to d. But the result of a careful consideration of this 
and many other points in machine design is that possibly such 
formulae as (4) are more likely to be misleading than useful. 
A complete theory to replace (4) is perfectly possible, but it has 
not yet had the services of the necessary good mathematician. 
"With (1) and (2) we feel much safer than with the others. 

7r 
Putting them equal, -rdy-^ = (p - d)tf^ .... (7). We need some 

other equation evidently. Now, in all design we try to obtain 
maximum advantage of some kind. To get maximum economy 
of material we were obviously right in trying to have a joint 
equally ready to break in various ways. 

But there is a more general kind of economy, not at all easy to 
express algebraically (see my Calculus for Engineers, Art. 37), 
which tells us to punch holes in thin plates up to f " thick and 
drill them in thick plates. For thin plates, then, we have such 
a rule as d^t^ or taking into account cost of riveting, say 
c? = 1^ + -^ if we are not to have too much expense in the 
fracture of punches [compressive strength of punch more than 
equal to shearing resistance of plate], whereas we have only ons 
of these things to consider in drilling holes. We may take the 
rule c? = 1.2 \/ ^ . . . . (8) as probably agreeing with the best 
practice from ^ = i to ^ = 1 inch. If this rule were followed, 

then, using (7) and (8), we find - 1-44 /s=: {jp — d) f^^ or 
f — c? = '^^T^fjft, =^ ^j say. The strength of the joint is the 

strength of the unhurt plate multiplied by ^- , or by 

P 

y .... (9) where a stands for p — d. For a rivet in double 

A -\- d ^ ^ ^ 

shear we use (2), (5), and (8), and find 2^ - cZ=-727r/'s//t, and we 
may call this A in the formula. It is just twice the last. 

ISTow for complicated kinds of joint we must make assump- 
tions which are less likely than the above. It is usual to 
calculate, instead oi p — d in (2), a width of strip i(; equivalen t 

in tensile strength to one rivet in single shear, or w=-r d~/Jt/^. 

Draw round each rivet a circle of diameter d + iv, and let lines 
come to the circles dividing the plate up into strips of the breadth 
w ; thus we allot a strip of plate to each rivet. Students ought 



APPLIED MECHANICS. 



339 



to scheme for themselves examples such as are figured in 

books on "Machine Design." In such books they will also find 

sketches of many riveted joints which are well worth study in 

case there are no actual specimens to look at. The result of 

the calculation is a formula like (9). 

The results of actual tests show that instead of A in the 

numerator of (9), we have a different number k A, and students 

Wiio have read Chap. XIII. carefully, and also what I have 

here given, will know fairly well why k is not equal to 

unity. We have, then, d = 1'2 V t, j) ~ K + d . . . . (10). 

k A 
Strength of joint == — , X strength of unhurt plate. . . . 

(11), where k is given in the following table : — 





Iron Plates. 


Steel Plates. 


Single Eiveted Drilled 


•88 


1^0 




Punched 


•77 


0-9 


Double „ 


Drilled 


•95 


1-06 


5> >> 


Punclied 


•85 


1-0 


Treble „ 


Drilled 




ro8 





Iron Plates and Iron 
Rivets. 


Steel Plates ai:d Steel 
Rivets. 




Drilled 
Holes. 


Punched 
Holes. 


Drilled 
Holes. 


Punched 
Holes. 


liap Joint or Butt ) Single Riveted 
Joint with. One Double „ 
Covering Plate ) Treble „ 


1-20 
2- 22 
3-23 


1-47 
2-66 


0-9 
1-7 
2-5 


1-08 
1-93 



All these values of A are to be doubled for butt joints with 
two covering plates. 

It is wise to slightly round the outside sharp edges of 
drilled holes. It is to be remembered that a punched hole is 
to be called a drilled hole if the plate has been annealed, or if 
the hole has been rhymered out after punching. 



340 APPLIED MECHANICS. 

The suitable joressure P lbs. per square incli of a boiler \» 

then p = t — — 7=/-^ R if ^ and R are thickne?s of plate 

and radius of boiler in inches, and / is the suitable tensile 
stress in pounds per square inch of the unhurt plate, and 
where a and k are the numbers gi^en in the above tables for 
the longitudinal seams. 

In the following exercises the working / is to be taken as 
10,000 lbs. per square inch for wrought ii'on, and 12,000 lbs. 
per square inch for mild steel. 

EXEECISES. 

1. Cylindric boiler 10 feet diameter, Ij-inch steel plates, with steel 
rivets, drilled holes ; butt-joint, with two covering plates ; treble riveted. 
What is the working pressure ? What are the pitch and the diameter of 
the rivets? Ans.^ p = 213 lbs. persq.m. p = 6-34 inches. ^=l-3i inch. 

2. Cylindric boiler 5 feet diameter, hon plates, iron rivets, double 
riveted butt-joint, one covering plate, drilled holes ; working pressure, 
150 lbs. per square inch. Find t. Ans., t = 0-67 inch. 

3. Find the dimensions and strength of a treble riveted butt-joint 
with double covering plates for plates | inch thick. Plates and rivets 
are of steel. Ans., d — 1-04 inch, p = 6-04 inches. Efhciency = '895. 

4. Determine the pitch of the rivets for a single riveted lap-joint of 
I inch plate, so that the joint may be equally strong to resist tearing and 
shearing. Diameter of rivets is | inch. Safe shearing strength 7,800 lbs. 
per square inch ; safe tensile strength is 10,000 lbs. per square inch. 

Ans., 1-813 inch. 

5. In a single riveted lap-joint the plates are |- inch thick, and the 
rivets | inch diameter ; calculate the pitch for the greatest strength of 
joint, the shearing resistance of the rivet being three-quarters of the 
tiensle resistance of the plate per square inch of section. 

Ans., p = 1-78 inch. 

6. The steel plates of a boiler are -^ inch thick, and connected by 
longitudinal double riveted butt-joints, with covering plates on each side ; 
find suitable proportions for a joint and calculate its efficiency. 

A)is., d = 0-9 inch, jj) = 4-3 inches. Efficiency = -79. 

7. In a single riveted lap-joint, assume that the shearing strength per 
unit of area of the rivets is equal to the tearing strength of plates per 
unit area. If the plates are % inch thick, diameter of rivets f inch, and 
the pitch 2^ inches, wiU the joint yield by tearing or by shearing? 
Calculate the efficiency of the joints. 

Ans., Tearing. Efficiency, -69. 

8. Two lengths of mild steel tie-rods, 7 inches x 1 inch, are to bo 
connected with double butt-straps. Determine dimensions and efficiency. 

Ans.. d = 1-3 inch. Three rivets each side. Straps, t = 0-647 inch. 
Efficiency, 0-81. 

287. Nature of Shear Strain.— When i/ moves to y, Fig. 178, the 
diagonal x y' becomes extended to x y. Imagine the angle y' » y^ 



APPLIED MECHANICS. 341 

as shoTSTi, to be an exaggeration of about one hundred times what 
it ever is in iron or steel. Its original length was \^'l (the 
diagonal of a square whose side is 1), and its new length is 

_ s 
a/2 4- — ^, as we see very easily. Hence the diagonal x y\ and 

all lines parallel to this diagonal, have a tensile strain, whose 

, . elongation ^„ * rr- i .-,- - s 

amount is . . ° -^ tv o^ /k -^ V^, and this is 77- Again, 

origmal length V2 ^ 2 »^'J^> 

in the same way we find that the diagonal y' x, and all lines 

parallel to it, have a compressive strain whose amount is — -. Thus 

2 

it has become quite clear that a shear strain sLmply coubists of a 
compressive strain in one direction, accompanied by a tensile 
strain in the perpendicular direction, these strains being each half 
the shear strain. 

We are led to speculate on the behaviour of material sub- 
jected to equal compressive and tensile stresses crossing inter- 
faces at right angles to one another. Consider a small right- 
angled prism of material, shown in Fig. 178, of which m f k. 
(Fig. 180) is a mag-nified cross-section. 
Neglect the weight of the prism, and 
for ease of calculation make m f, say, 
1 inch, M K the same, and let the length 
of the piism at right angles to the paper 
be also 1 inch. This prism is kept at 
rest by the matter outside it acting on 
its three faces. Face m f is pushed by 
a normal force of p' lb. per square inch, 
and as its area is just 1 square inch, 
the resultant push is p' lb. actiag at the 
centre of the square face m f. Similarly 

the face m k is pulled by a normal force of p' lb. And also 
the face f k is acted on by tangential forces of p lb. per square 
inch, and as its area is V2 square inch, the total amount of 
shearing force acting on f k: is ^ V2 lb. Now, when three forces 
keep a body in equilibrium, and two of them are at right angles, 
the smn of their squares is equal to the square of the third force 
(this is easily seen if we draw the triangle of forces) ; hence the 
square of pV2, which is 2p2, is equal to p''^ + p'^ or 2p''^. Hence 
p =^ p' ; and we have proved that the compressive and tensile 
stresses which occur in simple shear strain are numerically equal 
to what we called the shear stress. We have also to recollect that 
we cannot have shear stress parallel to the paper in planes 
at right angles to the paper without haAing equal shear 
stress parallel to the paper in a set of planes at right angles 
to the first and also to the paper (see Art. 282), 

Suppose we cut a cube abcd (Fig. 181) of 1 inch side, 
from a material subjected to simple shear strain, and let the 
faces of the cube parallel to the paper have, as before, no 




342 



APPLIED MECHANICS. 



stress upon them, the other faces being at right angles to the direc- 
tions of compression and extension. Shear occurs parallel to the face 

A c. Let us consider 
the motion of the point 

9 --:.;•.. I D relative to ac; in 

fact, regard a c as 
fixed. Under the sole 
action of the pushes 
on A D and b c we 
know that the side n c 
shortens by the small 
amount pa (see Art. 
265). Let us set this 
off from D to M. It is 
gTeatly magnified, as 
shown in Kg. 181. 
But when this occurs 
the side ad lengthens 
hy the amoimt p^ ; set 
this off from m to d'. 
Hence the pushing 
forces on ad and b c 
cause D to move to 
d'. Again, the pulling 
forces on D and a b 
f mother lengthen a d hy the distance ^;a, which we set off from d' 
to L, and shorten d c hy the distance ^^j8, which we set off from l 
to j>". Hence the motion of d due to the pulls and pushes acting 
together is d d", and we see that this is 

(d M + L d") ^/2 or (a -f /3) J) VT. 




1%. -81. 



V2 



-inch, 



But s, the amoimt of shear, is d d" -f- d o, and as d o 

as A D is 1 inch, we have 

s = (a + i8) pV2^ -^ or 1p (a + &). 

That is, shear strain =: shear stress multiplied by 2 [a + ^). So 
that the reciprocal of 2 (a + )8) is ivhat we called n, the modulus of 
rigidity of the material. 

289. General Results. — Eeferring hack to Ai-ts. 269 and 288, 
you will see that we have 

Modulus of rigidity n 



Modidus of elasticity of hulk ... b: = 
Young's modulus of elasticity ... e = 



2 (a + )8) 

1 

3 (a - 2)S 
1 



and you will also see that if we know two of these for any 
material, we can find the third. 

These resiilts are so important that we put them also thus: — 
shapes : — 



APPLIED MECHANICS. 343 

E 3n 9k 6n 9k' 3k+n' a 2n b* 

Some French, mathematicians have thonght that the ratio of j8 
to o (called Poisson's ratio, and always denoted by the letter a), 
and therefore the ratios of n, k, and e to one another, are constant 
for isotropic substances, a being always foiu- times i3. Experiment 
has shown that this is not the case, the ratio of a to /3 being 3 to 
2-5 in glass or brass, 3*3 in iron, 4-4 to 2-2 in copper, and in other 
substances Tarying from these values very much indeed. 

Just as Young's modulus is seldom found from experiments on 
the extension of wii-es, but rather from the bending of beams, so 
the modulus of rigidity is seldom found from experiments like that 
of Fig. 176, but rather from experiments on the torsion of rods or 
wires. 

In all the cases of simple shear which we have described we 
had more than simple compression and extension. There was a 
rotation of the lines in which the compression and extension took 
place. "When there is no rotation, and this is very easy to imagine 
(let A c not be imagined fixed in Fig. 181, but imagine the lines at 
45° to the horizontal to remain fixed in direction), so that the 
principal directions of strain, as they are called, remain the same in 
dii-ection, the strain is said to loej^tire (or irrotational), otherwise it 
is said to be " rotational." 

EXERCISES. 

1. For mild steel e = 30 x 10*5, n = 12 x 10^ ; find a, j8, and k. 

1 1 

^^'•' «=30 X 106' ^=^12 X 107' ^ = 20 X 106. 

2. The halves of a flange coupling are bolted together with six bolts 
at 6 inches fi'om the centre, 60 horse-power transmitted, 100 revolutions 
per minute ; safe shear stress, 8,000 lbs. per square inch. Find the jDroper 
diameter of each bolt. Ans., "41 inches. 

3. A 3-inch square steel bar, fixed at one end, loaded with 80 lbs, at 
its other end ; find the deflections due to shearing in the two cases ; 
length 3 inches, length 50 inches; according to Art. 281, and compare 
with the correct meth.od of Art. 369. 

4. Ten cubic inches of wrought iron and 10 cubic inches of water are 
subjected to fluid pressui-e of 3 tons per square inch; find the new 
volumes. If the iron is spherical, what are the old and new diameters ? 

Ans., Iron, 9-99664 cubic inches ; water, 9'79 cubic inches. 
Iron, diameters 2-673 and 2-6727 inches; water, 2*673 
and 2-671 inches. 

5. For wrought iron e = 29,000,000, k = 20,000,000 ; find a, fi, n, 
and calculate the value of Poisson's ratio. 

6. A cube of copper 3 inches edge is subjected to a hydrostatic pres- 
sure of 4 tons to the square inch ; find its new volume, k = 24,000,000. 

Ans., 26-99 cubic inches. 



344 APPLIED MECHANICS. 

7. For copper x = 5,600,000, e: = 24,000,000 ; calculate the value of e 
and Poisson"s ratio. ' Am., 15-6 x 10^ ; -385. 

8. A steel puncli | inch in diameter is employed to punch a hole in a 
plate f inch in thickness ; what will be the least pressure necessary to 
drive a punch through the plate when the shearing strength of the 
material is 35 tons per square inch? Ans., 51 -56 tons. 

9. In a fly press for punching holes in iron plates the two balls 
weigh 30 lbs. each, and are placed at a radius of 30 inches from the axis 
of the screw, the screw itseli having a pitch of 1 inch. What diameter 
of hole could be punched ''oy such a press in a WTL'ought-iron plate of 
^inch in thickness, the shearing strength of which is 22-5 tons per square 
inch ? Assume that the balls are revolving at the rate of 60 revolutions 
per minute when the punch comes into contact with the plate, and that 
the resistance is overcome in the first sixteenth of an inch of the thick- 
ness of the plate.* Ans., ri36 inch. 

290, We have been in the habit of testing material under tensile 
stress only, or compressive stress only, or shearing sti-ess only. 
Tests are now greatly required of the strength of material under 
combined tensile and compressive stress, these not being equal in 
amount as they are in shear. 

In an indirect manner ^\ e have evidence that if on an interface 

there is shear stress q and compressive 

stress p, the strength of the material in 

J i resisting fracture at this section by 

Jri shearing is as if a shear stress acted of 

the amount q — fxp, where /^ is a constant 

for the material. In fact, the compres- 

I sive stress strengthens the material in 

j / I its resistance to shearing. It is as if yu 

_a were a co-efficient as of friction, pre- 

I venting sliding. The evidence is the 

j fact that when strxits of cast Iron, 

j stone, brick, and cement are crushed, 

I fracture usually occurs at a section which 

; makes an angle greater than 45° with 

^'■ lao ^-^^ cross-section. 

^^^ "' In Eig. 182, A B is the cross-section 

of area a of a tie-bar or strut, and 
CD is a sloping section making an angle Q with the cross- 
section ; so that the area of c d is a sec. 6. An axial load w 

produces tensile or com]3ressive stress ^j = - on the cross-section, 

A 

and or 2h ^^s. in an axial direction on the sloping 

* 1 suppose that we shall always have academic exercises hke this and 
like Ex. 6 of Art. 172. The advanced student will notice (see Art. 486) that 
it is just as misleading as those absurd pile-diiver problems in which force is 
calciilated as the kinetic energy of the pile-driver divided by the distance 
through which the pile is driven, so that, if the pile is not driven any 
distance into the gi'ound, the force is infinite. In impact questions like these 
safety is only to be found in considering force as the time rate of transfer of 
momentuxa. 



APPLIED MECHANICS. 



345 



section. The stress on the sloping section consists, therefore, of 
a normal tensile or compressive stress p^ cos.^ 6 and a nhear stress 
Pi cos. d . sin. 6. The material does not tend to hreak at c 
D hy the normal stress if the material is isotropic, as p^ cos.2 e 
is less than p-^. The shear stress is a maximum, and equal to 
^^j if is 45^. Now, in materials like cast iron, stone, ■ brick, 
and cement, the ultimate shear stress is less than half the 
ultimate compressive stress, and it might he expected that the 
fracture of a strut would he at a section making an angle of 45° 
with the cross-section. But in every case we find the angle 
greater than 45°, heiug nearly constant for the same 
material. The assumption that there is a resistance to 
shearing of the nature of friction — that is, proportional to the 
normal compressive stress — is not only a reasonable - looking 
hypothesis, hut it agrees quite well with the observed 
phenomena. For a shear stress pi cos. 6 sin. 6 and a compres- 
sive stress pi COS. 2 e are hy our assumption to be replaced by a 
shear stress p^ (cos. 6 sin. — /x cos.^ 0) .... (1) ; and this is 

a maximum on the plane for which e = i6° -\-^ii ij. = tan. <p. . . (2 

2 

In cast iron, fracture usually occurs on a plane making 54|° with 

the cross-section. If om- hyxDothesis were reall}^ correct, our 

co-efficient of internal friction would therefore be -35 for cast iron. 

291. There is much published information on the fracture by 
compression of blocks of stone, cement, and bricks. In almost 
every case care is taken in loading the usually short specimens that 
friction at the ends shall prevent the material swelling laterally. 
When sheet-lead is inserted at the ends it gives a small amount of 
lateral freedom, and in every case the breaking load is lessened 
by its use, and therefore it is said to be wrong to use lead. 1 
consider all this published information to be nearly valueless, 
except that there is some proba- 
bility that half the usually pub- 
lished ultimate compressive strength 
for a cube is the true resistance to 
compression in the material. Hence, 
when the published strength of 
stone is given as between 250 and 
1.600 tons per square foot, we 
ought possibly to take it as truly 
from 125 to 800. In masonry 
structures the working stress is 
probably never as great as 50 tons, 
and generally it does not reach 10 
tons per square foot. 

292. To see to what extent 
we can, by means of lateral fluid 
pressure on a strut, strengthen it, 
we may provisionally use the theory 
above given. 

If there is compressive stress in a strut, p^, and a lateral 
normal pressure p2, and the material breaks hy shearing, let the 




Fig. 183. 



346 



APPLIED MECHAXICS. 

plane b c "be part of the cross-section, with, axial compressive stress 
acting through it ; let the plane a c have normnl stress ^^^ through 
it ; let the interface a b have normal compressive stress p and 
tangential stress q upon it ; consider the cquilihrium of a prism 
1 inch at right angles to the paper, and let a c = 1 inch. The 
■ normal pressure forces on the faces parallel to the paper balance 
one another independently. Equating the resolved parts of the 
forces vertically and horizontally, vre find 

p =Pi . cos.20 + jt?2 sin.20, 5- = (^j _ p^) COS.0 . sui.0. 
"We shall use the hypothesis that g will not produce fracture until 
it exceeds /hy the amoimt of the friction jxp; that is, we may take 
it that the value of the shear stress as a producer of rupture is 
only q — fxp, and it is easy to show that this is greatest when 

= 4 5 + ^....(1). This angle, therefore, is independent oipi 

and p.y Li q — fip is put equal to /g, we see that the least value of 
Pi to resist fracture is 

^ p^ (cos.0 sin.e — fi COS.-0) -/s 
^^ COS.0 sin.0 + fx sin.'^e ■ • - • [ ), 

or, indeed, we may say that the principal compressive stresses ^i 
and 2^2 ^^^^ produce fracture if 

Pi (cos. e sin. d — fi cos.^a) — Jh (c°s- ^ sin. 9 + ji sin.^ 6) >/s...{3), 
where 9 has the value given in (1). p^ ^^ supposed to he less than 
2^1. If 2^2 is a tensile stress, we have only to give it a negative 
sign in (3). (3) may be written nq^i — ')Uh>fs • • • • l'^)) "^^ere 
m and n are constants for the material. 

ISTow let ;;j, p^, and p he tensile stresses, and reverse the 
arrowhead on q- We have the same niunerical relations between 
Pi It Pv P-iy ^^"^ fractm-e occurs by shearing when /^ has the value 
q + fxp, so that fracture occm's when 

Pi (cos.0 sin.0 + fi cos.20) — p.2 {cos.6 sia.O — fi sin.^g) > f^, 
if is taken of such a value as to make q + /zj; a maximum; that 
is, if = 45° + ^/2 .... (2). In this 2^2 is supposed to be less 
than pi. li.2^2 is ^ compressive stress, we have only to give it a 
negative sig-n in (1). (1) may be written 7)1^2^1 — oi^pc^^f^, where 
m?- and n'^ are constants for the material. Let p.2 = 0. If p^ is 

compressive stress, = 45 + ^, and it will be found that 
-— = sec.(^ - tan.^ .... (4), 

o., = ti={i-(?^n/lf.,.,a,. 

If p-y is a tensile stress, = 45 — ?, and it will be found that 



t 
2 

Ml = sec.<^ + tan.v?) (6), 



APPLIED MECHANICS. 347 

using /c and /t as the usual crushing and tearing stresses of the 
material and eliminating <|) between. 

Given compressive stress pj^ and the value of / to find the least 

value of ^2 for resistance to fracture by shearing ; take = 45" + -^ 

as giving the plane on which the tendency to shear is greatest, 
whatever j}^ ''^^J 1^6, and we find 

- 2/s . cos. (p +pj{— sin. ^) 

•^^ 1 + sin.^ 

Thus, for example, taking cast iron, let us suppose that 
(j> = 28^ and ^u == 0-53, as akeady found, if the ultimate /^ = 
95,000, and fs = 28,500, then sin.t^ = '47, P2 = — 34,000 + . 36 p-^ 
(that is, p-^ increases 27,800 Ihs. per square inch for every increase 
of 10,000 lbs. per square inch in p^). 

If for any material f^, fp f^ are the numerical values (treated 
all as positive quantities) of the three stresses which the material 
will stand, 2/^ =/g (sec.(|) — tan.<^) = f^ {s,ec.(p + tan.^)). 

Thus, for example, ^ = 1+4^. 

Now in cast iron the proof-stresses given in Table XXII. are 

t = 21,000, /, = 10,500, /, = 8,000, -^ = 2 = 1jL!1^ and 

hence <^ = sin.-^ = 19°^, tan.(^ = -35, Oc = 5i% Ot = 35°^. 
With this value of (p, fg ought to be •356/^, whereas it really is 
•38/g. This is a discrepancy very allowable, and we may take it as 
some sort of verification of the theory. These were the first numbers 
I tried, but I have since found that other pablished numbers are 
less satisfactory in their support. For example, /J, ought to be 
greater than f^ for all materials if the theory is correct. It is 
evident that special experiments are required as a test. 

Tresca and Darwin have propounded the hypothesis that 
Pi — P2 is constant ; and it ought to be easy, by experiment, to 
decide whether this is constant, or whether, as by my theory, it 
increases as p-^ and p^ increase. For example, I make p-^ — p2 
equal to 94,000 when p^^ ^ in cast iron, and equal to 272,000 
when ^2 = 100,000. These are from the ultimate values. As to 
the permanent set strength of cast iron, I make j^^ — P^"^^ 21,000 
wheniPg = 0, and 121,000 when p,^ = 100,000. In fact^ instead of 
making ^^1 - i?2 = 94,000, I make_?;i — 2-78 p^ = 94,000. 

The hypothesis of Poncelet and St. Venant is that the material 
is fractm-ed when the strain exceeds a certain amount, and this 
hypothesis is often used to give ultimate strength conditions when 
the strains have been investigated mathematically. But in a 
wrought-ixon bar we have sometimes before fracture strains which 
are several hundi^eds of times as great as the greatest strains to 
which the calculations ajDply, and it seems to me, therefore, that 
Poncelet's theory has no probability of correctness in its favour. 

If we assmne that in any kind of earth there is a coefficient of 



348 



APPLIED MECHANICS. 



friction fx between layers, but no permanent resistance to true shear- 
ing (that is, /g=i: 0), we get Rankine's theory of earth pressure 
from the above theory. The lateral pressure p^ necessary to prevent 
a direct pressure ji^j from causing motion or fracture is given by 

p-2 1 — sin.d) 

i^i 1 + sin.^' 



Thus, if the static coefficient <>f friction in a 



1 __ .fiCQI 

certain kind of earth is '9, so that (p is 42°, then 2^.Jpi is .cmV 

or, say, \. "'' 

Example. — ^The weight of a building is 10'' lbs. The area of 
the concrete bottom of the fomidations is 2,000 square feet. At 
what depth ought it to be below the level of the soil, if the soil is 
such that (p = 4:2° ? 

Ans., the pressure pi lbs. per square foot is 5,000 lbs. Tc 
resist this, a horizontal pressure p^ of 1,000 lbs. is needed. 
Eegarding the horizontal pressure of 1.000 lbs. as a new p^, it 
needs a vertical pressiu-e of 200 lbs. per square foot, due to the 
weight of outside earth, to balance it. If the earth weighs 100 lbs. 
per cubic foot, the depth of the foundation must be at least 2 feel . 

Assuming that the theory is right, the difficulty in carrying 
out a rule of this kind is that we do not know </>, the angle of 
repose of a particular kind of earth. Rankine assumed that a long 
mound of earth would, after much weathering and rest, get to 
have a natural slope (p. If the natural surface of earth is hori- 
zontal, it is easy to find on the above theory the stress on any 
interface when motion is about to take place, and ]3articularly on 
a vertical interface, and so the reason for part of the following 
rule of Rankine's is known. The study of the stresses when 
the ground is sloping must be left as an exercise for students. 

293. Rankine's Rule for Earth. — Draw an angle x o r, Fig. 
184, to represent the static permanent angle of repose of the kind 
of earth. Describe Y r x, a semicircle touching or. If a b, 
Fig. 185, is the vertical face of a wall sustaining a bank of this 
earth, whose slope is A c, make the angle x p equal to the 
inclination of A c to the horizon. Find b d = o Q.A b/o p. Then 




Fig. 184. 




Fig. 185. 



A b D is a wedge of earth whose weight represents the total 
pressure acting on A B. The pressures act in directions parallel 



APPLIED MECHANICS. 



849 



to A C, and the resultant force, representing the total pressure, 
acts a third of the way up from b to a. 

This rule may be compared with the rule of Art. 173 for 
water pressure against a vertical wall. Rankine neglects fric- 
tion against the wall face. Students of this subject are directed 
to a paper by Sir B. Baker, Proc. Inst. C.E., Vol. 65, and its 
discussion. 

When grain is stored in vertical prismatic cylinders, the 
average pressure in pounds per square foot on the flat bottom is 
c dw where d is the diameter in feet, w the weight of a cubic 
foot, and c is 0-84 for wheat, 0-96 for peas. At greater heights 
than three times the breadth of section of a bin the pressure 
on the sides is constant, being about 50 lbs. per square 
foot for all kinds of grain until we get near the surface. 



294. Twisting^.— In Fig. 
186, AB represents a wire held 
firmly at A. At B there is a 
pulley fixed firmly to the 
wire, and this pulley is acted 
upon by two cords, which 
tend to turn it without 
moving its centre sideways. 
In fact, they act on the 
pulley with a turning mo- 
ment merely. But the pulley 
can only turn by giving a 
twist to the wire, and the 
amount of motion it gets 
tells us how much the twist 
is. A little pointer fastened 
at c moves over a cardboard 
dial, and tells us accurately 
how much twist is given 
to the wire. The angle 
turned through by the 
pointer is called the total 
angle of twist at c. If 
we had a pointer at each 
of the places g, h, and c, Pig. ise. 

and if A, G, h, and c were 
one foot apart from one another, we should find that 




the 



350 APPLIED MECHANICS. 

angles of twist at G, h, and c are as 1 : 2 : 3 ; in fact, the angle 
of tvnst is proportional to the length of wire tivisted. 

You will find that if a twisting moment of 10 pound-feet 
produces a twist of 4°> then a twisting moment of 20 pound- 
feet produces a twist of 8°, and, in fact, the twist is proportional 
to the twisting moment which is applied. You will also find 
that if you try dilBFerent sizes of wire of the same material, say 
wire whose diameters are in the proportion of 1, 2, 3, &c., and 
to each of them you apply the same twisting moment, the 
amount of twist produced in them will be in the proportion 1, 

— , — , &c. ] that is, inversely as the fourth power of the diameter 

of the wire. Lastly, taking wires all of the same diameters and 
lengths, but of different materials, and applying to them the 
same twisting moment, the amount of twist will he inversely 
projjortional to the number which we call the m^odulus of rigidity 
of the material. The exact rules are given in (1) and (2), and 
the values of n given in Table XX. may be relied upon in such 
calculations, because they have all been determined from 
experiments on the twisting of wires and shafts. 

Exercise. — A brass wire 20 inches long, O*! inch diameter, 
twists through a total angle of 130 degrees when a twisting 
moment of 4 pound-inches is applied. Find n for the material. 
Answer : — 3'6 X 10*^ lbs. per square inch. 

Exercise. — What would be the twist of a shaft of the same 
material with a twisting moment of 600 pound-inches, 20 feet 
long, 1*2 inch diameter ? Answer: — 7 '8 degrees. 

It is, however, well to notice that the drawn brass will 
probably have a different value for its N than the brass of a 
much larger shaft. 

It will be seen that the strain is a shear strain. Con- 
sider M H G (Fig. 187) to be a cross-section of the wire ; then 
a point which is at H before the twist 
occurs is found to be at G when there is 
a twist in the wire, and a point such as 
p' moves to P, but a point o in the centre 
of the Avire does not move. ISTow there is 
no such motion at the fixed place a. Fig. 
186, and in each section there is more of 
this motion the farther it is away from A ; 
in fact, the motion is just as it was in the 
Fig. 187. indiarubber of Fig. 176, only that it varies 




APPLIED MECHANICS. 351 

in the section, the motion being greatest at the outside 
of the wire, and nothing at the centre. The material 
breaks when the shear stress at the surface become* too 
great, and the rule found by experiment is that for any 
material, whatever the length of the wire, the twisting moment 
which will cause rupture is proportional to the cube of the 
diameter. It is well known that when a shaft is transmitting 
power, the horse-power transmitted is proportional to the 
twisting moment or torque in the shaft multiplied by the 
number of revolutions made by it per minute. The rule used 
by engineers is this :— c/= 3-3 {/n/n .... (1), as giving the 
safe diameter of a wroughtiron shaft at n revolutions per minute, 
if it is only subjected to torsion. Observe that if we double 
the speed, the shaft is strong enough for double the power. 
Instead of 3 3 we use 2-9 for a mild steel shaft and 4 for cast 
iron. 

294a. Shafts usually carry pulleys, and are otherwise loaded 
as beams, as by the pull of belts, and therefore, for reasons 
given in Art. 379, we take 1| to 1^ of the above size for mill 
shafting, and for crank shafts and shafts subjected to shocks 
we sometimes add 50 per cent, to the diameter as given by (1). 
We have some explanation in our theory of Art. 263 for this 
increase ; some of it is due to the variation in stress, and 
therefore to fatigue ; some of it is due to the fact that in 
crank shafts the maximum torque is often double the average 
torque. In a long line of shafting, if the power is given off at 
various places with some irregularity, it may even become 
evident to the eye that the shaft is perpetually twisting and 
untwisting, for of course the twist is proportional to the horse- 
power transmitted if the speed is constant. When this is the 
case, although the shaft may seem to be strong enough, it is 
weak because it is not stiff enough. A very long shaft some 
times gets into a state of torsional vibration just in the same 
way that the cage-rope of a coal-mine gets into a state ot 
longitudinal vibration. The nature of this vibration will 
depend on accidental causes, and should the impulses that give 
rise to it happen to repeat themselves at proper intervals, the 
vibration may go on increasing until the torsion at some place 
may be sufficient to produce rupture. In the same way a 
number of men walking from side to side of a large ship, just 
taking as much time in going from one side to the other as the 
ship takes to make a vibration, may make the rolling dangerous. 



352 APPLIED MECHANICS. 

it is for this reason that we endeavour to make the period of 
oscillation of a ship diflfer greatly from the probable period of 
waves which she may experience (see Art. 4 89). 

In very small shafting this vibration often occurs, and it is 
usual to add vaguely f to f inch to the diameters found by the 
above rule — a sacritice to the Goddess of Chance. 

295. Consider a little iDiism, p b (Fig. 188), Tvhose ends lie in two 
cross-sections of a shaft near together, o heing the centre of one of 
the sections, and o' the centre of the other. The twisting strain 
causes b to move to b', regarding p as fixed. (The motion is, of 

course, usually very much less 
, than I have here shown it). 
...'JO There must, then, he shearing 
,,--""' i forces acting on the ends in 
; opposite directions. If a is the 
, I angle of twist of the shaft per 
I inch of its length, then b o' b' is 
] a multiplied hy o o' ; and if o p 
j or o' B is r, then b b' is ;• ao o', 
I where a is an angle measured 
i in radians. The shear strain in 
: the little prism is bb' diAided 
...-- io hy PE or o o', so that it is ra; 
K --■;.'-■■'' j hence the shear stress is xra 
1 (see Art. 282). If « is the area 
of the end of the little j)rism in 
square inches, the shear force 
acting on it is N r a a, and as 
this acts in the direction at 
^■?- 1^^- right angles to the radius, its 

moment ahout o o' is n" r'^ a a. 
But we have a similar moment for every such little area into' which 
the cross-section may he divided, and to find the total torque we must 
take the sum of aU such terms. Now, n and a are the same every- 
where, so that in taking such a sum our only difiiculty is with the 
factors r^a. But the sum of all such terms as r^a is called the 
moment of inertia of the section about the axis o o', and it has heen 
calculated for us. Thus, if d in'tho diameter of a roimd shaft, the 
moment of inertia of its section ahout an axis thi-ough its centime 
ai rigJit angles to the section is ic y>^ -j- 32, and for a hollow shaft 
whose outside diameter is d and inside diameter d, the moment of 
inertia is tt (d^ — d^) -^ 32 ; and henco we see that the twisting 
aioment t necessary to produce a twist of a radians per inch in a 
round shaft of diameter n is t zzzirs a 0^32 .... (1), and for a 
hollow shaft it is t = tt n a (d* — <?^)/32 .... (2). The torsional 
rigidity of a shaft is defined as a if a = t/a. The values of A will 
he found in Table XV., Art. 532, for various sections of shaft. The 
verification of these rules is an cxcc41ent laboratory exercise. 

296. The strength of a shaft is to be calculated on the assump- 
tion that rupture occurs when the shear stress N?'a mentioned 




APPLIED MECHANICS. 



353 



above exceeds the greatest shearing- stress to whicli the material 
ought to be subjected ; and as the stress is greatest when r is the 
outer radius of the shaft or | d, so that / = |^ n d o, and as from 
equation (1) (Art. 295) we find that n o is 32" t 4- itd'*, we know 
that/=: ^D X 32t -f- ttd^, and this is the condition of strength 
of a cylindric shaft. It is more compactly put in the form 
7rD3/ 



16 



for solid cylindric shafts .... (1), 



and in the same way we get 

T =: — — -^^ — for hollow cylindric shafts .... (2), 

/ being the breaking shear stress of the material in pounds per 
square inch, t the twisting moment in pound-inches which will 
cause rupture, d the outer diameter, and d the inner diameter (if 
the shaft is hollow) in inches. 

We see, then, that the strength of a solid shaft depends on the 
cube of its diameter, whereas its stiffness depends on the fourth 
power of its diameter. 

As to the practical rule given in (1) (Art. 294) , we saw in Art. 182 
that torque in -powo-di-feet, multiplied by angular velocity in radians 
per minute, divided by 33,000, is the horse-power. As we use t in 
pound-inches, t == 12 x 33,000 x H/27r«. If we use this in (1) of 
this article and take/:= 9,000 lbs. per square inch, we find the usual 
practical rule (1) (Art. 294). Taking /= 4,500 for cast iron and 
13,500 for steel, we find the rules for these materials already given. 

297. Shafts Subjected to Twisting and Bending. — It is most 
convenient here to assume that a student has read the larger print 
in the next chapter, and knows that if a round shaft of diameter 
D inches is subjected to a bending moment m pound-inches there 
are compressive and tensile stresses through the cross-section at the 
circumference of amount 2^ = ?>2 -is.! k n^ = 2aM, say .... (1) for 
a solid shaft, and 32 m d/tt (d^ —d^) = 2 o m, say .... (2) for a 
hollow shaft, if ^ is the internal diameter. We have just found 
that the shear stress on the same interface, due to a twisting 
moment t, is /==aTand5T. 
shafts. The other stresses across 
other interfaces at the point 
ought now to be studied, but it 
will be found that it is only 
necessary to determine the prin- 
cipal stresses there — that is, the 
greatest and least tensile or com- 
pressive stresses which act across 
any interfaces. 

298. In Art. 292 we had a 
simple case of the general rule 
for stresses given in Art. 290, 
and this is another case nearly 
as simple. 

Suj)pose that across an inter- 
face A c at right angles to the 



. . (3) for the solid and hollow 




»»» 



354 APPLIED MECHANICS. 

paj^er we kuoAv that tliere is a normal stress (either tensile or 
compressive) jj lbs. per square inch and a shear stress /; that on 
planes parallel to the paper there is no stress ; that on the planes 
at right angles to ac and the paper there is no normal sti-ess. 
Consider the equilibrium of the prism a c d, and find the traction 
on A D of (7 lbs. per square inch. For the sake of ease of calculation 
we "will assume a c to be 1 inch, and that the prism is 1 inch at 
right angles to the paper. Kow, when there is shear stress / 
across a c there is also an equal shear stress / across d c (Ai't. 282). 
AVhat are the forces with which stuff outside acts on the prism ? 
The resultants of the forces on the faces are : — On a c, a horizontal 
force 2^ and a vertical force /, because a c is 1 square inch in area ; 
on DC, a horizontal force / x area of n c or /. d c or /. cot. 6, 
as A c is 1 ; on a d, a force q x area of a d or g . cosec. 6, and this 
force and its direction are what we desire to calculate. Given 9, 
it would be easy to calculate q and its direction ; but our problem 
is even simpler. It is this : Find on what plane a d there is only 
a normal stress q with no tangential component, and find q. 
Eesohdng forces horizontally, / cot. 9 -[- p z=: q cosec. 6 x sin. 9, 
or / cot. 0+ !> = ?.. ..(I). Eesolving forces vertically, 
f-=zq. cosec. 9 . cos. 9, or f = q . cot. 9 . . . , (2). Hence, as 
(2) gives ns cot9:=flq, using this in {1), we find f'^Jq + 2^=iq or 
q^-pq =/2 ; so that q = lp ± ^.^WTP • • • • (3). 

It is easy to find 9. There are two answers, differing by a 
right angle. A stress is called a principal stress if it is normal to 
the interface, and we see that we have in (3) the two values of the 
principal stress due to a combination of tensile 2^ ^'^^ shearing 
stress /, such as we have suj^posed. The principal stresses are 
across interfaces which are at right angles to one another. 

Example. — At an interface ^^ z= 6 tons per square inch and 
f z=. b tons per square inch, then $' =r 3 + 5-83 ; so that the 
principal stresses are 8 •83 tons per squai-e inch in tension and 
2-83 tons per square inch in compression at right angles to one 
another. 

Exercise. — Wrought ii'on is not to receive a greater tensile or 
comjiressive stress than 5 tons jjer square inch. There is already a 
tensile stress of 4 tons per square inch across an interface. AVhat 
shear stress may also cross that interface ? 

Ans.., ^j =r 4 and <? = 5 = 2 + ^^4 + p fi-om (3), and hence 
/= 2-24 tons per sqiiare inch. - 

Exercise. — A roiuid shaft is in torsion, and the shear stress 
produced across the section near the circu mf erence is 8,000 lbs. per 
square inch. At the same section the shaft is subjected to bending, 
and a compressive stress of 6,000 lbs. per square inch is produced 
across the same interface. What is the greatest compressive stress 
in the material there ? 

Ans., q = 3,000 + V9 x 10^ + 64 x lO^ =: 11,544 lbs. per 
square inch. 

Exercise.— Jl p> and / are equal, find them that q may be just 
6 tons per square inch. 

Ans., Each of them is 3*71 tons per square inch. 



APPLIED MECHANICS. 



355 



299. We see, then, tliat a hollow round shaft subjected to the 
twisting moment t and the bending moment m is really subjected, 
at points near the circumference, to the maximum compressiye or 
tensile stress q where <? = Jm + -J W- ^ -\- h'^ iP- (see (2) and (3) of 
Art. 297), or 



16d 



(d^ - d'^) 



M + a/ M^ +T 



(4), 



and in solid shafts 



16 



q = , \ M 



,/ y-i + t2 



. . (5). 



Hence the greatest 

existing in a hollow 

T and M is exactly the same as the 

gi^eatest shearing stress in a shaft 

subjected only to a twisting moment 



and least compressive or tensile 

or solid shaft when it is subjected to 



M + v/ M- + t2 



(6). 



A . 








^ 


rr 


~" r 


— 


'J i i 


"v 




~lc" 


\ 


1 

-rt 













300. An overhung crank shaft (Fig. 

190) is subjected to a wrench. If the 
gTeatest force exerted at the pin a, at 
right angles to the crank, is e, the twist- 
ing moment is F . A c, and we may take 
the bending moment as r . b c if b is the 
middle of the j ournal. Hence the wrench 
is equivalent to a twisting moment 
F (bc+ a/bc^ + Ac2), orF . (b c + ab), 
a rule which is perha]3S a little un- 
expected. 

301. Ordinary shafts are of -^-rought iron or mild steel, materials 
such that their resistances to compression, tension, and shearing 
are not very different, and therefore when they are subjected to t 
and M we at once calculate (6), and say that the strength of the 
shaft is the same as if it were subjected to this tvidsting moment 
only. The practical rule (Art. 294«) is worked on the idea of a twist- 
ing moment only, om d <j:\/ t. If there is also a bending moment 
M^ which is of the value 'k t, then the rule becomes e'xddentlv 



Fig. 190; 



(? = 3-3 y M+^/ 



V M-^ + t"^, d 



3-3 Vt . V Ti + x/Z;2+ i. 



The exti-eme values of h for many kinds of shafting have been 
worked out, and the conseqiient change in the multiplier of (1), 
by means of shrewd guessing and calculation ; but to my mind 
it is a better recommendation of the rules given in Art. 294 « 
that they are consistent with theory and with the best practice of 
engineers. 

I have given here the usual theory of shafts subjected to 
twisting and bending ; it assumes that the strength is determined 
by the maximum stress. A true theory would take account of the 
considerations of Art. 294a. 

If a tensile stress y and a shearing stress / act on the same 



356 



APPLIED MECHANICS. 



interface, and if }?■ and ^^^ ^re the jDrincipal stresses, we see from 
(3) of Art. 298 that 

Ih = lP + ^\Jt±Jl'> 

Ih = \V - ^^P'' + /'• 

The theory of Ai'ts. 290 and 292 tells us that fracture will take 

place if vi^pi — ri^jh > fsi where ni^, n^, and/g are constants which 

ought to be known for the material. If j; is a compressive stress, 

we have fracture if 7npi — njj^ > fg- 

The probable values of m, n, vi^, n^ for cast iron are given in 
Art. 292. For wrought iron and mild steel it is possible that cp of 



Art. 292 is 0, and hence m = J, n 



= h, n^ = h. Hence 



■we have fracture either on the compressive or tensile side of a 
shaft — that is, whether p is compressive or tensile stress — if 
2 \/ ^p^ +f'^ >/a • Hence, if a shaft is subjected to the twisting 
moment t and tbe bending moment m, we calculate an equivalent 
twdsting moment a/t- + m'-^, and assume that the shaft is subjected 
to this alone. 

For materials in general, and probably even in the case of 
wrought iron and mild steel, the equivalent twisting moment 

ought to b e calculat ed from 
hu + k a/t^ + M'^, where 
h and A* are constants, depend- 
ing upon the nature of the 
material, which are not 
known at th(! present tinie. 





302. The demonstration of Art. 296 is found to agree with experi- 
ment, but its results must not be applied except to shafts w^hich are 
circular in section. Our assumption, which experience warranted, 
was that when such a shaft as a b (Fig. 191) is fixed at b, and when 




Fig. 1D3. 



APPLIED MECHANICS, 



357 




Fig. 194. 



to an arm, cd, a twisting couple is applied, every straight line 

in a section remains straight, and moves through the same angle 

as every other line. But it can he shown that this is not the case 

for a shaft of any other than a circular section. Thus, let o (Fig. 

192) he the centre of gra^dty of 

the section pqs, and let us 

suppose that a shaft of this 

section is subjected to the sort 

of strain I have descrihed. The 

shear strain at the point p is 

in the direction p K, perpen- 
dicular to o p. Let its amount, 

which we know to he o p x 

angle of twist, he represented 

by the length of p k. It is 

easy to show that this is just 

the same as a shear strain p n 

in the direction p n, normal to 

the surface of the shaft at p, 

together with a shear strain in 

the direction p t, tangential to 

the shaft at p. But shear 

strain in any direction is al- 
ways accompanied hy a similar strain in a plane at right angles 

to this direction (see Art. 282), so that since we have the shear 

p N, we must also have a shear parallel to the axis of the prism 

along the surface at p, and this cannot be produced merely by 

a twisting moment. We must 
imagine that along with the 
tmsting moment there is a 
force distributed over the sur- 
face of the shaft to produce 
the above effects. The result 
of an exact in^^estigation (Art. 
313) is that a twisting couple 
produces a gTeater twist than 
might appear from what I have 
said in Art. 295, and it also 

produces a warping of the naturally plane sections of the shaft. 

Thus Fig. 193 is the shape assumed by each section of an elliptic 

shaft, and Fig. 194 of a square shaft. Imagine a 

section to be distinguishable, say, in a glass shaft 

by a thin layer of a different colom- from the 

rest. Deeper shading indicates greater distance 

from the observer who is looking towards the 

fixed end of the shaft. The arrows show the 

direction of the twisting moment. In the fol- 
lowing three sections, instead of the torque for a 

twist of one radian being equal to n times the 

moment of inertia of the cross- section, it is only 

•84 times this for a square section (Fig. 195), -54 times it for the 

section Fig. 196, and '6 times it for the section Fig. 197. Indeed, 




Fig. 195. 



Fig. 196. 




Fig. 107. 



358 APPLIED MECHAyiCS. 

the square section has only -SS times the torsional rigidity of a 
cylindric shaft of the same sectional area : Fig. 196 has -67 times, 
and Fig. 197 has -73 times the torsional rigidity of a cyHndric shaft 
of the same sectional area. 

Then-omliersinthe column headed w. Table XT. , express the rela- 
tive strengths to resist twisting of the various sections there figured. 
The torsional rigidity of an elliptic section whose principal 
semi-diameters are a and h is s k a^l^j[a- + iP'). If m is the 
twisting moment, the shear stress at a point x. y (the axis of x 
heing a) is 2 M ^/h^x- + o^y-l-KdHr^. This is greatest at the end of 
the minor axis, l»eing 2 ujiralr^. 

The torsional rigidity of a rectangle, if its length is two or 
more times its breadth, is, with some accuracy, the same as that 
of the inscribed ellipse multiplied by the ratio of their poLar 
moments of inertia. The greatest shear stress occurs at the middle 
of the longer side of the boundary, and is 3 m (a- -j- Z^)/8 a^b-, if h 
is the twisting moment and a is half the longer, h half the shorter, 
side of the rectangle. 

303. A very interesting result of the investigation is that there 
is always greatest distortion at that part of the surface of a 
shaft where the sui-face is nearest the axis. Thus in an elliptic 
shaft the substance is most strained at the ends of the shorter 
diameter of the section. Imagine a very light box to be made 
so as to contain frictionless liquid exactly of the shape of a 
shaft. If we give a sudden turn to the box about the axis, 
the liquid will be left behind if the box is circular in sec- 
tion, but it will have motions relatively to the box which can 
Tery readily be imagined if the shaft is not circular in section. 

^, Xow the actual velocity of 

the liquid at any place re- 
latively to the box is in 
the same direction as, and is 
proportional to. the shear in 
a similar shaft when it is 
twisted. This has been 
proved by Lord Kelvin. You will see from 
this that there is very little strain at the projecting ribs of the shaft, 
whose section is shown in Fig. 196, and just at the projecting 
angles of Figs. 195 and 197. This reminds me of a general remarfj 
which I have to make, and which I must leave without proof. A 
solid of any elastic substance cannot experience any finite stress 
or strain in the neighbourhood of a project- 
ing point, unless acted on by outside forces 
just at the point. In the neighbourhood of 
an edge it may have strain only in the direc- 
tion of the edge, and generally there will be 
exceedingly great strain, and stresses at any 
re-entrant edges or angles. An important 
application of the last part of the statement 
is the well-known practical rule that every 
Fig. 200. re-entering edge or angle ought to be rounded 

to prevent risk of rupture in solid pieces 





APPLIED iTECHAyiCS. 359 

designed to bear stress. An Ulastration of the principle is the 

stress at the centre of the drctLlar outline in the three sections 
of shafts Tig-5. l&S, 199, and 200 . In Tm. 19S. at o. there is 
no stress ^hen the shaft is twisted; in Fig. 199 the stress may he 
calcTilated : in Fig. 200 the stress is exeeedmgly great for exen the 
smallest twist (see Art. 302). 

EXEECISES. 

X.B. — ^A most nsual error of students is to forget that momenta in 
pound-inches are not numerically the same as in pounJ-feet. Beginners 
had better leave the exercises involving bending moment until they have 
studied bending. 

1. A shaft 1 inch in diameter can safely transmit a torque of 2,400 
pound -inches. What diameter of shaft would be required for transmitting 
l.'t H.P. at 200 revolutions p>er minute Y Aus.. Ij inch. 

2. Find the horse-power which may be transmitted by a shaft 4 
inches in diameter when running at 150 revolutions per minute, if the 
stress due to twisting be limited to 9.000 lbs. per square inch. 

An$., 269. 

3. A line of steel shafting is 8 J feet long: if a twisting moment of 
4:,Cm1»<3 p-jund-inches be applied at one end. what will be the total angle of 
twist, the 'iiameter of the shaft being 2i inches r AVhat horse- j«ower 
will this transmit at 220 rev.jlutions per minute r Atis., 4-7' ; 14 BLP. 

4. A solid wrought-iron shaft is to be replaced by a hollow steel shaft 
of the same diameter. If the material of the latter is 30 per cent. 
stronger than that of the former, what must be the ratio of internal to 
external diameter ? "What is the j>ercentage saving in weight ? 

Aug.. -693 : 4S per cent. 

5. The amount of twist in a solid shaft is to be limite<i to l"" for each 
10 feet of length. Find the diameter for a twisting moment of 50 ton- 
inches, the modulus of torsional rigidity being 10,CMJ/.K>D lbs. j>er square 
inch. Aug., 52 inches. 

6. A wrought-iron shaft is subjected simultaneously to a bending 
moment of 8,000 pound-inches, and to a twisting moment of lo.OOO 
pound-inches. Find the twisting moment equivalent to these two and 
the least safe diameter of the shaft, the safe shear stress l:»eing tak^i at 
8,000 lbs. per square inch. Ans.. 25.000 pound-inches: 252 inch^. 

7. Find the diameter of a shaft for a winding drum which works 
under the following comiitions : the Load lifted is 1^ ton ; diameter of 
drum. 5 feet : width of face of drum, 26 inches : distance from inner fac-e 
of drum to the middle of the bearing of shaft, 13 inches: maximijm 
stress, 7,000 lbs. per square inch. Ans.. •5-44 inches. 

8. A wrought-iron shaft 3 inches in diameter, and TnaVirig - 140 rerola- 
tions per minute, is supp'orte-i at wall brackets 16 feet ar-irt. There is a 
pulley on the shaft, midway between the l-earings: if "he resultant side 
pull due to the weight of the pulley and the piil of the c-elt be 210 lbs., 
what is the greatest horse-power the shaft will transmit with safety : Safe 
shear stress, 7,800 lbs. per square inch. Ajis., 66. 

9. A bar of iron is at the same time under a direct tensile stress of 
5,000 lbs. per square inch, and to a shearing stress of 3.500 lbs. per 
square inch. What would be tiie resultant equivalent tensile stress on 
the material? Am., 6, SGI lbs. squHre inch. 



360 APPLIED MECHANICS. 

10. Taking the safe tensile stress of wrought-iron to be 10,000 lbs. 
per square inch, determine whether it would be safe to subject a piece of 
wrought-iron to a tensile stress of 3^ tons per square inch, together with 
a shear stress of 3 tons per square inch. 

Ans. Unsafe ; max. stress = 11,700 lbs. square inch. 

11. A wrought-iron shaft is subjected to a twisting moment of 36,000 
pound-inches and a bending moment of 18,000 pound-inches ; find the 
diameter when the maximiun shear stress is 8,000 lbs. per square inch. 
Find also the twisting moment which alone would produce a shear stress 
of the same numerical value. Ans., 3-3" ; 58,200 pound-inches. 

12. A screw propeller-shaft 10 inches in diameter is subjected to a 
twisting moment of 35 ton-feet, and to a bending moment of 10 ton-feet, 
due to the weight of the shaft and the pitching of the ship. What is the 
maximum compressive stress if the thrust of the screw be 10 tons ? 

Ans., 2-9 tons. 

13. A shaft 12 inches diameter, transmitting a twisting moment of 
100 ton-feet, is also subject to a bending moment of 20 ton-feet. Find 
the maximum stress induced. Ans., 4*3 tons per square inch. 

14. Find the diameter of a wrought-iron shaft to transmit 90 horse- 
power at 130 revolutions per minute. If there is a bending moment 
equal to the twisting moment, what ought to be the diameter ? 

Ans!., 2-7 inches; 4-1 inches. 

15. A steam-engine crank is 12 inches long, and the greatest force 
which is transmitted through the connecting-rod is 9,000 lbs. Find the 
diameter of the wrought-iron crank- shaft, taking the safe shear stress at 
9,000, the distance of the centre of the crank-pin from the centre of the 
bearing nearest it being 10 inches, measured horizontally. 

Ans., 5-07 inches. 

16. A round shaft 3 inches diameter, find the sizes of equivalent 
shafts of square, elliptic and rectangular sections if the breadth and 
thickness of each of these latter are as 1 to 2. If these shafts are 20 feet 
long, and they are transmitting 20 H.P. at 100 revolutions, what is the 
total twist on each of them, n = 10,500,000. 

Ans., Eectangle, 2-15 x 4-3 inches; ellipse, minor axis, 2*38 inches; 
square, 2*73 inches side. Twist on circular shaft = 2'073°, on square 
shaft, 1-78°; eUiptical shaft, 1-05°; rectangular shaft -93°. 



361 
CHAPTER XV. 

MORE DIFFICULT THEORY. 

304. Mathematicians endeavour to help engineers (including 
in this term all men who apply the principles of natural science) hy 
investigations concerning ideal elastic material shaped like actual 
%eams and shafts. The mathematical analysis is exact and 
difficult, and only a few problems have yet been solved, and it is 
only by leaving out terms that seem insignificant that we are able 
to apply the results to actual problems. The engineer recognises 
from the begimiing that his problems are too complicated for any 
exact mathematical in^-estigation. He therefore leaves out his 
apparently insignificant terms rather at the beginning than the 
end ; but indeed he leaves them out in any part of his investiga- 
tion if they are likely to give trouble, for he recognises from the 
begiiming that his theory is only to guide him, and that the final 
appeal must be to experiment. The engineer looks upon the 
phenomena involved in the loading of the tie-bar as simple because 
experiment is easy; whereas the mathematician, seeing that a 
lateral contraction accomxDanies an axial elongation, regards it as 
complicated. 

The engineer ought to study, and develop, and correct, by 
experimental observations, his usual method of investigation as 
described in this book, but he ought also to study the mathema- 
tician's treatment of the subject, and let it assist him as he lets 
experiment assist him. The following very short sketch needs 
much time for its proper comprehension ; it is from the mathe- 
matician's point of view. Students may pursue the subject in Mr. 
Love's treatise on elasticity, or Thomson and Tait's treatise on 
natural philosophy. 

305. The consideration of homogeneous strain in general (when 
any portion of stuff in the shape of a sphere is changed into what is 
called the strain ellipsoid, any three co-orthogonal diameters of the 
sphere becoming conjugate diameters of the ellipsoid, planes 
parallel to one another remain parallel ; all parallel lines get the 
same fractional changes in length) is not so difficult as it is 
tedious. Unfortunately the authors of 

books insist on its being studied. For 
our purposes we have only to deal with 
infinitesimal strains, and this is easy. 
Suppose that a point a*, y, z is dis- 
placed to X -{■ ti, y -{- t\ z ^ w where 
w, t7, tv are very small, then 

du J. dv dw ,,. 

' = dx^^^dy^'j;='iz'---^^\ 

are the tensile strains of the stuff in 
directions parallel to x, y, z. In Fig. 
201 let the axis of x be at right angles Fig. 201. 

M* 




362 APPLIED MECHANICS. 

to th.e plane of the paper. Let the traces of two planes, c b and c a, 
parallel to o y and o z, and perpendicular to the paper.hecome changed 
to cf b', c' a'. The angle by which a' c' b' is less than the original 
right angle is evidently a shear strain. I shall call the amount of 
it a ; and as it is a shear of planes normal to y, parallel to z, or a 
shear of planes normal to z, parallel to y, there is no great harm in 
calling it a shear ahout the axis of x. jN'o'sv the angle turned 
through hy c a, clockwise, is really the horizontal motion of a minus 
the horizontal motion of c, divided hy ac, since the angle is 
very small. But this is (Ivjdz. Let the student be quite 
sure of this fact. Similarly the angle, anti- clockwise, tiu-ned 

through by c b, is dwjdi/, and hence (^ = -j h -^ • • • • (2). 

Similarly if b and c are to y and z what a is to re, then 
, da , dw dv , du ,^. 

A student may keep these in mind by means of the mnemonic 

a b c \ 

X y z\ . . . . (2). 



u V w 



V 



Notice that the average rotation of c a and c b, clock's^'ise, in Fig. 
201, is half the algebraic sum of the angles tiu'ned through by ca 
and c b, or, calling the average rotations of the material w^ about 
the axis of x, ih.j about the axis of y, d-^ about the axis of z, 

T Idw dv\ ^ , fdtc dw\ ^ . fdv du\ ,„, 

It is evident that when w^ = wo = wg = 0, or there is no rotation^ 
it means that the strain is pure, or that the three principal 
diameters of the sk-ain ellipsoid, called the principal axes of strain, 
remain parallel to their original directions. . 

Shear strain involves no change of volume ; and if the edges 
parallel to x, y, z of the imitcube become 1 + e, 1 -f /, 1 + <7, the 
volumetric dilatation or strain is e -\- f + g, since these are small. 
Or if T> is used for this, then 

du dv . dw 

The conditions Jj = Wg = wg = are evidently the conditions 
that there is a function (p (called the strain potential), such that 

d<b d<p d(p , 2^ j_ • ji • 

u r= -,--, V = -J-, zc = ^- ; and we see that, m case there is no 
d.T dy dz ' 

cubical dilatation, (4) becomes 

d-cp d-2cp d-2<p _ 

d^2 + J^+-J^2-^-"- (^)- 

At any point o of a body let there be three planes of reference 
meeting at the three mutually perpendicular axes of x, of y, and of z. 
Let tensile stress be called positive. Across the plane of yz (usually 



APPLIED MECHANICS. 



363 



called the plane x, tecause tlie axis of x is normal to it) let the 
stress he resolved into its three components — Xa;, parallel to ox; 
Yx, parallel to oy; and zx, parallel to oz. Across the planes i/ 
and ;:; let the component stresses he Xy, Yy, Zy^ and x^, y^, Zz, 
To find the stress components f, g, h in the three directions across 
any plane whose direction cosines (or the direction cosines of its 
normal) are I, m, n, consider the equilihrimn of the tetrahedron 
formed by the four planes under the action of the tractions acting 
from material outside it. The area of the sloping face being a, 
the areas of the other faces are I a, m a, and n a. The resultant 
force on each area is stress multiplied by each area. We have, 
then, resolving parallel to x, FArrxaj^A + x^/mA + XaWA 
and two other equations. Hence, 



F = Z xa? + w xy + « X2 
G = l YX + m Yy + n yz 
n = l zx + m zy + n zz. 



P3 = iP31+yP32 + /.P33- 



In a fluid the static stress is always normal to any interface. 
Hence f, g, and h are in this case the components of a normal stress 
p on the new plane, and all the tangential stresses vanish. Hence, 

F=:Zp=:Zxa; or P =: Xa; 
Q =zm-pz=mYy or p-=:Yy 
nzzin-p zrinzz or pr=Z3; 

80 that the stress is the same across any interface whatsoever. 

306. Now consider in the general case the equilibrium of a paral- 
lelepiped whose corner is at o (Fig. 202), the co-ordinates of which 
are x, y, z, and let the co-ordinates 
of the opposite corner o' be 
a; + Sa?, 7/ + Sy, z -f Sc, and let 
us consider the equilibrium of aU 
the surface tractions acting on the 
faces from the outside. The re- 
sultants of the normal forces meet 
in the centre, and if we neglect 
volumetric forces, they are in 
equilibrium. They are not shown 
in Fig. 202. 

We show the tangential forces 
per unit area with which outside 
stuff acts on inside on three of 
the faces. The other three faces 
have similar forces, the arrows 
being in opposite senses to these.* 
Hence their moments are just the 
same as the moments of these about axes through the centre. 




Fiff. 20-2 



* It is easy to take into account the fact that there may be volumetric 
forces, and that z a; on the x plane through o' is not equal to z on the x 

plane through 0, but is rather Zx-{-^x — Zx. We do this later, and it is 

easy to see that these extra terms vanish in the present problem. 



364 



APPLIED MECHANICS. 



Taking- moments of the figured forces about the axis through 
parallel to x, we find 

Zy . dx . 5z . ^di/ = Yz^x . d^/ . ^5z; so that Zy = Yz. 

This is Cauchy's theorem. We call each of them s. In the same 
way we have the other two relations hero giA^en : — 

Zy =Yz ^^ S; Za;=X3=T; Xy =Yx ='u. 

We also give the names p to x a; q to y y, r to z 2. 

s, T, u are the shear stresses, and v, q, and k are the normal 
tensile stresses in the material. Hence our equation may be 
written : — 



G = / u + w? Q + w s [ 
-R= It + VIS + « R ) 



.. (1). 



Exercise. — Across what interfaces at a j)oint are the stresses 
normal ? That is, find the principal stresses and their directions. 

If r, G, and h are the components of a force b normal to the 
plane I m n, then b ^ = f, Bm =z g, b w =: h ; and if we sub- 



stitute these in (1), remembering that l^ + m^ 



= 1, we 



can find lirm and b. The answer tells us that there are always 

three directions and amounts of princi2)al stress. 

In our most general state of stress, Fig. 203 shows the surface 

ti^actions acting on the element ^x S>/ Szfrom the outside. Imagine 

equal and opposite forces on the 
other three faces. 

Now let us consider volumetric 
forces and the rates of variation of 
the stress. On the faces meeting at 
o we have p q r s t and v. But 
the forces on the other faces are, as 
regards normal forces, 




p + 8a? 



dx' 



Q + 5y 



R + 8z 



^R 



clij' 



On 5y . Sz we have also m -{■ Zx ,— 



and 



(I T 

5a; . -^ 
ax 



on Sz . Ix we 



Fis. 203. 



on 8a; . 8v we have t + 82; 



have u + 8v .-^ and s -f 82 . -^- 
dy dz 



d T 



+ 



d s 
ih' 



NoAv we know 



that the parts of these forces p and p, s and s, etc., balance in- 
dependently ; therefore we need only consider the extra ones as 
shown in Fig. 204. 

If the "^^olumetric force in the direction of increasing x is 
p . yi . Zx . 'by . Zz, then 



APPLIED MECHANICS. 



365 



fj'S. 



Bu . Sz + 5i/ . Sz . Sx ~ + hj 



+ 



d T 

dz' 



5y 



dy 



.dx.dz 



0. 



"We have similar equations for the other dii'ections, and these 
reduce to the following equations of equilihrium : — 



^, , dT . dv L^T ^ 

dx dy dz 

dx d\j dz 

dx dy dz 



(2). 



Note that we have 
on the houndary 
of the element to 
halance, in prov- 
ing Can chy's the- 
orem, or that the 
bodily forccshaA^e 
no moment. They 
may have some 
moment in the 
matter in which 
magnetic 
stresses act, but 
in ordinary stress 
phenomena we 
assume the 
truth of Cauchy's 
theorem. 

If the mate- 
rial is fluid, so 

that S = T = U=:0, 

dv 



assumed the 



forces 



p X + 



dx ~ ^' 



p Y + 
p Z + 



0, 



= 0. 




do. 

d^ 

dz 
Example. — Let 
p X be called %o, • Fig. 204. 

the force on unit 

volume, say the weight in pounds of 1 cubic foot, then p is pressure 
xn pounds per square foot at any depth in a fluid under parallel 
vertical s-raAdtational force. 



If there is motion in the general case, we must use x — 



d'^u 



instead of x merely, where u is the displacement in the direction 



366 APPLIED MECHAXICS. 

of increasing x of the mass p . 5.r Sy Sz ; and hence we have three 

equations hke _- + __ +- +px = p— , .... (3). 
dx dy dz "^ df- 

At any element of surface of a hody {I m n heirtg the direction 
cosines of its normal), if there are surface tractions with com- 
ponents r G H, the condition (1) is to be satisfied at the siu-face. 

Now, taking the unit cube, and letting a be — , the reciprocal 

of Young's modulus, and letting /8 be the lateral contraction in a 
tie-bar of the material, when a is the axial elongation ; imder the 
action of p, q, and e, the edges become lengthened by the amounts 
<?, /, and g, where ^ = p a — (q + r) )3, or 

e =1 Po — Q)3-Ri8) 

/— _P^ + Qa - K^ (4). 

g=i -P/8 - Qj8 + Ea) 
Compare these -^-ith Art. 269. 
Solring these equations, we find, writing d for e + / + <7, and 

recollecting the fact that a = — +— =-, ^=-1 — i-, 
^ 3n 9k e 6n 9k' 

a=:AD + 2n/ .... (5), 

where A stands for k - -| n ; k being the modulus of elasticity of 
bulk, N being the modulus of rigidity. 

If s, T, and u are the shear stresses, and a, b, and e are the shear 
strains (see Ai't. 305), s = x a, t = n J, u = n c . . . . (5). In 

that Art. we used -^—, etc., instead of e, etc., and so we can write 
dx 



die 
dx 



, r. dv 
AD + 2n 



i\--^^^' 



where d stands for -y- + ^- + -— , and we also have 
dx ^ dy ^ dz ' 

rZ w = v . de + q, . df -\- -b. . dg -\- s, . da + i . dh + u . ^c. . . (6) 

was proved by Lord Kelvin to be a complete differential in two 
cases — first, at constant temperatiu-e ; second, when the changes of 
sti-ain occur so quickly that no heat is gained or lost by the stufit. 

Using (5), we find that <^ w is a complete differential if 
2 ^' = (A + 2 At) (^ + / + ^)- + M («2 + ^2 + c2 _ 4/^ -^ge 

-4^/).... (7); 

^- , d\i ^ d\f , _^ 

so that p = —5- , etc., u — -3—, etc (7). 

dc de ^ ' 



APPLIED MECHANICS. 367 

Kirchhoff pointed out that w cannot be negative. When w is 
the whole body mores as a rigid body. He also showed that if the 
six strains be given at every point in a body it is necessary to 
impose six iadependent conditions, such as those of St. Venant 
given in (7) of Art. 308. Without these there may be " rigid 
body " displacements as well as the strains. He also showed that 
if the surface displacements or surface forces on a body are given, 
there is only one solution possible (Kelvin showed that there was 
also an unstable solution). Kelvin had shown that one vvas alwaj's 
possible. 

St. Venant showed that in calculating the str.dns due to surface 
loads it is only in the neighbourhood of the place that the actual 
distribution of the surface force is of any importance. This is 
called the priuciple of equipollent loads. 

Converting the stresses in (3) into strains, we have 

^^ + ""^ ^"^ + "" • ^'" + ^ "^ ^ p';rl • • • • ^^^' 

^2 ^2 ^2 

with two other equations, v^ means -j-g- + -j-^ + Tr '-^^^ 

surface conditions become r = ^ (A i) + 2 n e) + w n c -}- m n 5, with 
two others. Translating this, we have the surface conditions (1) 
becoming 



H:=wAd + 2nI-^j + l(a2~- m&i \ 



.... (9). 



Here dn^ is an element of the normal to the surface. 
(8) or (3) may be written 

with two other equations. 

If we neglect x, y, z, the volumetric forces, differentiate (10), 
etc., with regard to x, etc., and add, we get 

(A + 2n)v2d = p||....(11), 

a well-known equation showing that there is wave propagation 

with velocity = \/ • Differentiate the third of (8) with 

respect to y, and the second with regard to z, and subtract, and 
we get 

NvV'i = p-ri---.(l2), 



368 APPLIED MECHANICS. 

and two other equations. These are equations of distortional wave 

propa^tion '""ith a relocity = a / . 

307. To illustrate this method of study, let us consider the prob- 
lem of the thick cylinder cf Axt. 275. There we considered it from 
the stress point of view. "We shall now consider it from the strain 
point of view. At any point in the material at the distance r from 
the axis there is a radial displacement, which we shall call w, as if 
the direction r were oxir old direction x. There is displacement at 
right angles to u. and its fractional amount — that is, the strain — is 
evidently u/r, because if a radius r increases to r -\- u, the circmn- 
ference of the circle increases to 2 ir (r -i- u), so that the fractional 
increase of the circumference is u/r. Let us imagiae any tensile 
strain parallel to the axis, which we call the direction of z, to be 
constant. It is evident, also, that the strain is irrotational, and also 
that the three principal shears are 0. We make,these statements to 
show that we are considering the old practical problem ; but the 
mathematician puts it rather in this way : he assumes certain 
strains to exist ; he works out a mathematical problem, and he takes 
his chance of the results fitting any practical case. 

Let -p = e. - =f, g = q,-,, a ^ 0, b== 0,c = 0. 

1. No bodily forces, and the problem a static one — that is, the 
strains independent of time. Then t> = —^ + -, and equation (8) 
or (10) becomes 



<^ + -)i(^ + 7) = °----w- 



Hence 

^ + l!^'_4 = 0....{2). 

And we find on trial that '^ = a /• -f b r~^ . . . . (3) where a and b 
are arbitrary constants. EJnowing the strains, equation (5) (Art. 
306) enables us to find the stresses. 

p = 2 (A -i- x) A - 2 >• B /•""-.... (4), 
Q = 2 (A + n) A + 2 N B r-2 . . . . (5), 
11 = AA + 2N<7f, . . , . (6). 

The first two of these are the values obtained in Art. 27-5. 

Exercise. — Show that d is constant throughout the shell. (2) gives 

a-i( r- H - 

+ dr = 0, 

dr^ o 



which, is 



The first intesrral is 



APPLIED MECHANICS. 369 



= 0. 



^ 

d,'^ + 


r 


dti, u 




*^ + r 


= C( 



constant. 



But D = -A — , and is therefore constant throughout the space for 
dr r 

which the differential equation (2) holds. 

2. Rotating Cylinder. — Volumetric force aPr acting- radially on 
unit mass, a 'being angular velocity. Using this in (8) or (10) of 
Art. 306, instead of (1), we have 

<^ + ^»);|.(S + -:) + ^--°-'--'^)- 

If we try u = a )-^\ we find that n = l,n=- I will satisfy (7) if 
A is an' arbitrary constant ; also n = 3 will satisfy (7) if a = 
— pa^/S (A + 2 n) ; and hence 

u = Ar-\-Br-'^- pa^r^jS (A + 2 n) (8) 

where a and b are arbitrary constants. Having the strains, we 
can calculate the stresses 

/die u\ du 

=^ = HI^ + ?) + ^^T?■••■<"' 

(du , n\ , ^ n ,,-, 

^ /du , ?/A ,,,v 

If we let p = when r = r^, and p =^ when r ^r^, and we let 
a = 0, we get the answers given in Ai-t. 275 for a thick cylinder. 

Taking \\]) this rotational case by the method in the text (Art. 
275), in the case of a cylindric body rotating with angular velocity 
a, if p is the mass per unit volume, taking into account the 
centrifugal force on the element whose equilibrimn is considered, 
equation (1) becomes ^j . dr -\- r . dp - r'^pc^ . dr = q . dr, and 

this is 2> -\r '^'-j-, — r'^po? = q. The solution now becomes 

^ = c + Br-2 + ipaV (12). 

Let us take j? = 0, both outside and inside. 

0^c + Brr^ + ipa%2, 
0--C + Bro-^ + ipa%^. 



370 APPLIED MECHANICS. 

From these -we find 



1 „«2 I ^-r - ^-n'^ , _2 ;. 2 \ _ 1 o ro -2ri2- r,-2;-,2 
^P« l;.^-2-,^-2-'0 2-^, )-,pa ^ ,.^_2 _ ,.^-2 '> 

This is greatest where r = 7"p, 

To apply oiir results to a rotating solid cylinder, since we hare 
no statement of the inside stress, the central condition heing that 
there is no radial displacement of the material at the centre, it is 
necessary to write our equations in terms of the strains rather than 
the stresses. Using, then, the equations of Art. 306, letting ?-^ = 0, 

u = A)- -\- nr-i — ° ^ — - ?'3, and hence b = if the cylinder 
b (A + 2 N) 

is complete to the centre. We know, therefore, 21 everywhere,, 
and we can, if we please, now use the values of p and q. 

The solution is only to be regarded correct for a cylinder, not 
for a thin disc, because the above gives a value for k, the stress 
parallel to the axis, and there is no surface force on the disc. For 
the disc we use Dr. Chree's more correct solution. Dr. Chree's 

solution is this : — Let 

We can now easily write out the equations. Try if the solution is 

n=^{l-cr)[{3 + a) ah'- (l + cr) r^ } +^0- (l+o") r {P-3z^, 

tv==~fPcr\{3^(T) aH-^ (1 + a) r^z [ _ ^ ^2 i±f ^ (^2 _ ^2). 

it is easy to get the stresses from these values. 

In this solution the planes 2 = ± ? are free from stress, and 
there is no tangential stress on the circumference, and the resultant 
normal traction per unit length of the circimiference is when 



dv 


^ u dw , du dw. 


U' 


-^=r' ^^r/.' ^=.Zz+^ 



^j: 



r = a (that is, l v . dz = when ?■ =^- a). By the principle of 

equipollent loads, this means that there is no practical departtire 
from the real condition of things beyond a short distance inside 
the edge. 

308. Bending or Twisting a Prism.— A straight prismatic sur- 
face with plane ends bounds isotropic material. Surface tractions 
are applied to the ends only. The axis of the prism is the axis 
of z, one end being the origin. When we speak of bending, the 
deflection will be in the direction of x (that is, bending will take 
place in the plane xz). 



APPLIED MECHANICS. 



371 



Tte student will refer back to Fig. 203 to see the exact significa- 

du 



tions of the stresses p, a, R, and s, t, u, and the strains e = 



/ = 



dy' 



dw 



and a = 



dio dv , 
d^^dz'^ 



du 



dw 
dx' 



~ dx^ 



dx' 
du 

dy\ 



also to the equations of equilibrium (3), and to the equations (5) 
which express the stresses in terms of the strains, and also to the 
boundary conditions (1). He had better write out these equations 
here. We do so, and give them new reference numbers. We 
assume neither Tolumetric forces nor dynamical forces in (3), now 
caU.ed (1). 



d-p dv dT _ 

dx dy dz 

dv do, ds 

dx dy dz 

dr I <^s dn 

dx dv dz . 



.(1). 



e = p a — /8 (a + k) 
/=Qa-)8 (P + 11) 
^ = R a — )8 (p + Q) 



• (2), or 



.(2). 



Q = A D + 2 n/ / 

R = AD + 2n^ I 

S=Na, T = N^, U=Ne/ 

D stands f or 6 + / + ^. A is k — | n. Our old 
^ 6 N 9 3l' a 



3n"^9k 



(Poisson's ratio) , k is the modulus of elasticity of bulk ; n is the 
modulus of rigidity, and the surface tractions are 



V = Iv ■\- mv -\- nr 
G = ^ U + Mi Q, + « s 
H = Z T + ?n s + 7i R 



(3). 



It is to be remembered that the' mathematician assumes a 
certain condition of strain, and takes his chance of its fitting some 
particular practical problem. If it happens to fit the conditions of 
the problem exactly, it is an exact solution of the problem ; and, 
according to Art. 306, it is the only solution. If it does not exactly 
fit the conditions (as, for example, Dr. Chree's solution of the 
rotating disc), we discuss the discrepancy to see whether it is 
essential or negligible for practical engineering purposes. 

St. Venant studies the case of the prism under the following 
assumptions : p = a = u = 0..,.(4). There is no normal 
stress, therefore, in the dii-ections x ox y (that is, laterally from 



372 APPLIED MECHAXICS. 

fibre to fibre); there may be tangential force acting from fibre 
to fibre in a direction parallel to z. Hence equations (1) give 

dz dz a 3^ oij iiz ^ ' 

The surface conditions (3) become 

/t + ms = . . . . (6). 

He also a-sumes that the origin is fixed, and that particles in the 
directions of the axes of z and y just there are fixed. This means 
that at the origin 

,. = 0,. = 0,er = 0/:^ = 0. ^' = 0, ^ = 0. . . . (7). 
dz dz dy ^ ' 

The following purely mathematical work is tedious, but quite 
easy. The student will do it carefully for himself, following our 
directions. Inserting (4) in (2), so as to calculate the stitiina, 
we find that 

dn _ dv _ dw 

di-lTj--''Tz--"^^^ 

where a is Poisson's ratio, or yS/a, or ^A./(a. -{- n). Also r = 
m.earLS that 



-^- = means that --^ v + -^ 5- = 0, / 

dz dz- dz . d^- f (10^ 



^s , , - (l-w d-v 

-^— = means that -^ ^ — f- ^-^ = u 1 

dz dz . dy dz- > 

and the last condition of (5) gives 

/ d^u . d^ip , d^w . ^^ \ . d^u . dH 



a-w a-w a^ \ ani 



\dx . dz dx- dy- dy . dz) d-r . dz dy . dt 

+ (X+.2K)g = 0....(ll). 

The first and second of (10) and (8) give 

d-u Id-u d^v Id^ . . 

1^ = adx^^^dF^ =-aW " - ^ ^' 

(11) and (S) fifive, after simplification. 

Differentiate (13) with respect to z, and (10) with respect to j; 
and y, and use (8) to eliminate u and i\ and we get 

5 = 0. ...(U). 

Differentiate (10) with respect to y and x, add and use (9), and 
we eet 

^!L^=0....(i«). 
dx .dy . dz 



APPLIED MECHANICS. 373 

BifiEerentiate (10) with, respect to ij and x, and use (8) and (14), 
and we get 



dx'^ . dz dif' . dz 

Hence -- is linear in z, and linear in x and y separately. Hence 

^=a + a,x + a,?/ + z {^ + ^,x + ^,y) .... (17). 

[a and ^8 are here any constants, and not the coefl&cients of formula 
(2)]. Now use (8) with (17), and also (9), and we hare 

U = — (T {ax -\- I ai«2 4- a2yz -\- fizx -j- fi^V^^^ + i ^i^'^z) — 
^aiz^ — fpz^ + zUi-\- Vq 

^Yhere i(j and Uq are functions of y. Similarly, 

«;= — 0- (a// + a^^xy + i a.^y- + /3r^ + fiixyz -\- ^ l3.-,yh) — 
i a^^" - i ^2^3 _|_ ^.^ _^ ^,^^, 

where v^ and Vj are fimctions of x. Now try these in (9), and find 
values of ?/q, ■2/^, Vq, i\, and get 

to = z {a + a^x + a2.y) + 2" ^^ (^8 + fi^x + /3.?/) + (p {x, y) zz 0. 
Using this in (13), we see that we must solve 

To get the complete solution, we find the general solution wlien 
the right hand side is zero, and add to it a particular solution. 
For the particular solution, try 

(p^=zAx^-\-By^-\-c xy^ + D x^y, 

and we find it to be 

This, then, is what we must add to the general solution of 

-7-^ + -^ = 0, taking care that the value of (b so foimd enables 
dx^ dy ° ^ 

the whole value of w to satisfy the boimdary conditions. We 
may now write out w, 

W^z{a.+ aiX+ a^y) +^^2(^8 + ^l^ + ^^y) - | { i )8 (a?2 + y2) 

309. Boundary Condition. — Use (2), and find from the above 
values of v, v, iv values of s, t, and r. Use them in (6). IntegTate 
bcLh sides of the equation round the boundary of a cross-section, and 



374 



APPLIED MECHAXICS. 



transform the line integrals to siuface integrals over the section. 
This gives us )3 = 0. Hence our complete answers are : — 

u = - <r {ax + h a-^X' + a^ry + I fi^x-z -t- PnXyz) 

V = - <r (ay + a.xy + i a^y^ + P^xyz + ^ ^..fz) v ,,g. 

~ ^cuf^ - ^P.;f^ + iaa.>x'' - fi,xz + iafi.yX^z /^ • • • • V ; 

w = z {a + ajZ + cu,)/) + irz- {fi^x + fi.,y) + (f) - 

\xy {^y + ^■>>:) 

where d) satisfies -^ + -y% = at all points of any cross-section, 
dx^ dy 

and the condition 
7^4> , .., '^<P 



^;^ + "^^ = ^°^'"^-^^^ + 



^•2 h>nay^- + (| " '')^^^'^ + i/«a;2) -»i<r:r2j .... (19) 

at all points of the cylindric boundary. The stresses at any point 
are p = 0, q = 0, r = 0. 



hy'^ 



d<p 



%-^^-i^-^)^^y-i^^ 



(^y 



Q-.). -.„.)] 



(20). 



ii. = B[a + a-^x + a^y +z {fi-^x + ^8,2/)] 

By giving any values to o, a^, 0-2, /Bo- fii, P2 we get all the possible 
solutions. It mu5t be remembered that a point xy z has the new 

position X -\- u, y + v, ic + z. 

d(p 



^N'otice that if p^, jS^, fi.j are 0, then <f> = 0, because s = x 



and T = X -^, and each of these is at the boundary, and the only 



dx 



d(p d<p 



conditions to be satisfied by <b are that ^ ^ + »i -^ = at the 
•^ ^ dx dy 

boundary. Hence <^ may be a constant, or rather a function of z ; 
but it is where ; = 0, and hence (f) = everywhere. 

310. Example 1. — Let all the arbitrary constants vanish except a. 
We must remember that a.\s, any arbitrary constant, and is not the 
o of Art. 289. Then u = - trax, v = - aay, tc = oz, k = e a, and 
the other stresses vanish. "SVe have, therefore, a tie bar. 

311. Example 2. — Bending. — Let all vanish butaj. then ^^ = - 
\ ttj (z- + ax- - (T7J-), V = - a-^a-xy, v: = a^xz. Hence p = 0, q = 0, 
R = E a^x, s = 0, T = 0, u = 0. The shape of the centre line is 
obtained by putting a; = 0, y = 0, and we have u = - ^ aiz\ t; = 0, 



APPLIED MECHANICS. 375 

to = 0. So that it becomes the arc of a parabola, or nearly a ckcle 

The total force parallel to 



Si 



z across the section = I I e a^x . dx . di/ = 0, because x is measured 

from the centre of area. Hence the tractions at a section are a 
torque merely, called here a bending moment, whose amount is 
evidently - e ai i^ where i^ is the moment of inertia of the section 
about the axis of y in the section. Since w = a^xz, and we make 
z = Z(^ a, constant, we have the displacement at all points in a 
section. The section remains plane, and turns through the 
angle z^a^. 

Change of Shape of Section. — Take a portion of the section 
of rectangular shape. The boundaries, ha^dng been x=^a,y=±b, 
have become x ■= ±a - ^a^a- {a^ - y'^), y = ± b - a-a-^bx. So that 
the boundary consists now of two straight lines and two arcs of 
parabolas. We note that the upper and lower surfaces of the 
beam are anticlastic stu-faces, the principal curvatui^es being in the 
ratio 0". Note (Art. 330) the result of the ordinary theory. It is 
then evident that— at all events, when the bending moment is 
constant — the hypothesis on which our ordinary theory is based, 
namely, that a plane cross-section remains plane, agrees with the 
more fundamentally correct theory. But we have now to notice 
that this case of the St. Venant theory only contemplates small 
displacements, and may not be applied to cases where the displace- 
ments are large. The ordinary theory is more useful, therefore, 
for \^ makes the shape of the centre line to be circular, as it 
obviously must be ; and no attention need be paid to the fact that 
this case of the St. Yenant theory makes it the arc of a parabola. 

312. Exam^ple 3. — Let all the constants except fi^ vanish, and let 
)8o be called - t. Then u = - t zy, v = t zx, tv = (p {x, y), where 
T is a constant and w is a function of x and y only. "We find that 



and 



P = a = B = 0, s = X (I + T.r), T = K (g - ry), V = 0, 

at the boundary Ij^ + m~- = r {Jy - mx). Everywhere in the 
section -^ + -^ = 0. The total tangential force parallel to x is 



T . dx dy. On wiiting this out, it will be found transformable 

dx . ay^ 



"•"■IlBi'frr-")!**!-!!*")!] 

and into n I \lx {-?- ~ '^y) -^ '"-^ \^ "^ '^ ^) \^^ round the 
boundary, and this is by the boundary condition Hence the 



TNI + 



376 APPLIED MECHANICS. 

resultant force parallel to x is 0, and the resultant force parallel 
to y is 0. So that the forces over the section are a torque 

+ T x\ - y \-Y~ ~ '^p) [<^^ • (^y, which is equal to 

N I \{xY--^J-Y■\dx.d1/ = qNrI, say, where i is the 

moment of inertia of the cross-section about its centre. 

This case, then, is that of a prism subjected merely to a 
twisting moment ^ n t i. When the section is circular, ^7 is 1, and 
T is the angle of twist per unit length of the prism. When the 
section is not circular, q is not 1, and is called St. Tenant's torsion 
factor. To solve the problem, then, when the shape of section is 

given, we must find (p such that --^ + '-,-% — ^ ^^^ *^^'^^' ^^® section, 

with the boundary condition I -— -{- ni-^ — t {ly - mx). To do 
this, it is well, as in many other such problems, to find, first, a 
coniiioate funotio]i i|/, such that ^r4 + —~ = 0, and — = — ^, and 

-r- = - -,-. So that the boundary condition becomes 
dy dx ^ 

^d-y-'^'i^'"^^'^-'"''^- 
Now we see by trial that if ;// = i t (^^ + y'^) + c, the boundary 

condition is satisfied for -~ — t y. and -—■ = t x. It is evident, 
dy dx 

then, that if we can find a solution of -r^ + -~^ = 0, subject to 

Cl'iA/'^ CI (J 

the condition ;f/ = i t (a;^ + y'^) + c at all points of the boimdary, 
we have the correct answer. 

Example. — Prism of elliptic section, ~ +72 ~ ^' giving the 

shape of the boundary. It is found by guessing that t|/ = a (a;^ - y^) 
is a solution, if a is properly evaluated to satisfy the boundary 
condition. So that a {x^ - y-) — ^r {x^ + y-) + const, at the 

boimdary — that is, where y'^ = — («2 _ ^2^^ y^Q substitute this, 

and find that a is ^t (a^ - b'^)/{a'^ + V^). So that ^ = a [x'^ - y'^), 
and <p — - lAxy. 

Evidently the curves in the section along Avhich (^ is constant 
are rectangular hyperbolas, with the principal diameters as axes. 
The total twisting couple is 



# 



M = TNI+ 2 AN I \{- X"- + y')dx ,dy = /^2 ^ ^2^2 ^'^^- 



APPLIED MECHANICS. 377 

The shears are now easily calculated. 

S = n(- 2Aar + Ta;), t = n(- l^y - Ty\ 
( d^ - }?\ / 2 W- \ 

/ (P- - W-\ / 2 ^2 \ 

S is greatest when x = a^ then 

S = N T -R rs, also S^ + T 2 = /^ = — \ O^X' -\- fl^ 

or + b^ ^ [a- + b~)~ [ 

The maximum stress anywhere in the section is the value 

of T when y = b, viz. nt -^ p. For / is a maximum when 

^4^2 ^ ^4y2 ig greatest. Now, whatever be the value of x for 
which the maximum occurs, for that value of x the maximum will 
be when y is greatest — namely, on the boundary. Put x = a cos.0, 
y = b sin.0, then b'^x'^ + ahf- becomes aW- \}P- + («2 _ V^) sin.^0], 
which is obviously greatest when Q = 90°. Hence the stress is a 
maximum at the extremities of the minor axis. 

Example. — Torsion of a Shaft whose Section is an Equilateral 
Triangle. — Let 3 a be the vertical height of the triangle, then the 
equation of the boundary is [x - a) [x - y -v/^ + 2«) (a; + ?/ a/ 3 + 
2«) r= 0, or a;2 - 'ixy'^ + 3a (a;^ + ?/2) - ^d^ = 0. The function ^ = 
A {x^ - S xy'^) satisfies the differential equation, and a {x^ - 3 xy'^) 

— 5 T {x^ + y-) will be constant over the boundary if a = - — . 

Hence <if = - Z.^ - Zxy% and <|, = - ^ (^3 _ 3^2^^) .... (i). 

From this it is easy to find the strains and stresses. It will be 
found that the shear is in the corners, and is a maximum at the 
middle of the sides. 

313. Approximate Formulse. — St. Venant worked out and the 
strains and stresses for a number of sections, and he found that if 
the section of a shaft is not too different in any two of its 
dimensions across the centre, the torsional rigidity (or twisting 

moment per unit angle of twist) is a = ?n — where s is the area 

of the section and i its moment of inertia about its centre of 
gravity ; n is the modulus of rigidity, and m is a number which 
does not greatly differ in different cases, m = -02533 for an 
ellipse; and in the sections examined its lowest value was -023, 

and its highest was '026. Consequently, if we take a = — s^n/i in 
all such cases, there is no great error. 



378 APPLIED MECHANICS. 

314. Non-uniform Flexure. — In Art. 309 let all the constanta 
be zero except fi^. The displacements are 

V = - a-^jXi/z, 

w = cp + fi,\hz^x - xf-^ 

where (b satisfies -^ f — ^ =r at all points of a normal section ; 
d.c~ di}- 

and at the homidary the condition 

The stresses at any j)oint of a section z = constant, are 
s = y> i-~ - (2 + 0-) fiixi/ \, parallel to the axis y, 

T = X ( ~ - I ^^ {(TX- -f 2 y-) V parallel to the axis x, 
R = E /8j^z.r, parallel to the axis z. 

The total force parallel to x (really the shearing force x at the 
section) turns out to be e i jSj, and the resultant total force parallel 



is E )8i I I xy . 



to y (call it a shearing force y) is e jS^ I 1 xy . dx . dy, and this is 



if the axes are j)rincipal axes ; and we shall assume them to be so. 
The resultant stress parallel to z vanishes. The couple about the 
axis of y is - zfiiEJ. ■ The couple about the axis of 2 may be 
written out. By a combination of this and previous solutions, we 
have the case of any twisting and bending couples applied to a 
prism. If we merely take the case of a prism being bent, being 
fixed at one end, and loaded at the other with a load w, we must 
make our twist ; and hence we must use the solution in which a^ 
was constant, together with this in which ySj is constant, so that 
the two twists may just neutralise each other. We get this 
condition, we find, by taking a^ and yS^, such that cj + fi^l — 0. la 
this case we have a total tangential force parallel to x of the 
amount w = x = EiSii, and a couple EySji {I - z), or w {I - z), 
which we usually call a bending moment, due to a load w at the 
distance I from the fixed end of the beam, ai = - ^J = - w //e i. 

The equation to the centre line is x = — (i~^^- i^^)- We 

see, therefore, that the ordinary theory is right as to the shape of 
such a beam. The curvature aj is the bending moment at the 



APPLIED MECHANICS. 



379 



fixed end dhdded by e i, the flexural rigidity. Tlie displacements 
are, if there is no twist, 



V = —a- (I - z) xy, 



w = 9^[('' - 2/0) X - 2x1/] + cp. 



Notice from tt and v that the change of shape of the section is 
as o-iven in Art. 311 



The strain g =~=:~{z - T) x (that 



dz 



is, the tensile strain at any point of the section is proportional 



to x). 



da w 

~ - e = --^ (T [l 

dx E I 



~-)^,/- 



[I - z) X. AVe may 



write out the shear strains a and b in terms of <p, etc. c = 0. 

Now in cases that have heen considered in which x and y are 
small compared with I - z,\t is formd that cp is of the third degree 
in X and y. So that a and h are small compared with e, f, and g. 
For the same reason, the term 2 xy'^ is not important in iv above. 
And the engineer's theory based on the assumption that a plane 
section remains plane, may be taken as correct. When x and y 
are not small compared with I — z, we must find cp, and this is 
difficult. As 10 contains cp, although the tensile strain is pro- 
portional to X, the plane section does not remain plane if beams are 
short. St. Venant's solution assumes that w is distributed in a 
particular mamrer over the end section. But by the principle of 
equipollent loads the actual distribution of w is of very little 
consequence, except close to the end section itself, and hence is of 
no practical importance except in beams that are not long in 
comparison with the 
values of x and y in 
their sections. 

315. Vertical loads 
are often applied to 
beams on their hori- 
zontal top surfaces. 
We know from the 
principle of equipollent 
loads that the actual 
distribution has little 
effect except in the 
neighbourhood of the 
surfaces to which the 
loads are applied. We 
can obtain a fairly 
clear notion of th-e 
effect by thinking of 
the load on a plane surface bounding an infinite elastic solid. 




?ig. 205. 



380 APPLIED MECHANICS. 

M. Boussinesq has given the solution.* The following hrief 
memorandum may be useful, o o being the plane surface 
boimding the infinite isotropic solid, m is a point within, 
situated at a distance mk = 2 below the surface, k is any 
element of the surface, situated at the distance k m = ?• 
from the point M, and subject to a given exterior surface 
pressure k^^ = p per unit area, having the component kj;^ — ]?^ 
per unit area along elm. The pressiire which a plane ele- 
ment E E^ taken through m parallel to the surface o o will 
support per unit area in consequence of the pressm^e iJ will be 

3 h^z 

found directed along k m produced, and is equal to m f = 3 . 

If, as a particular case, the pressiu'e k p = ji? be normal, then 
l)^ — p cos. NMK =p -, and mf = 3pz-j2Tr?''^. If we want the 

vertical component of m f, we have Zpz^j^ irr^. 

Generally. — Let Wi be the normal force per unit area at any 
point on the plane bounding surface at the point x = x^, y = y'^, 
2 = 0, the axes of x and y being in the plane, and the axis of 2 
being the normal to the plane drawn into the material. 



<^ = I \ii\. rdx^ . dy^, and x = I I u\ 



Let <^ = 1 I ?^'i . '''^^'^ • dy^, and x = l I u\ log. [z + ?•) dx'^ . dy 



where r is the distance from .r^^O to xyz. Then u, v, and w being 
the displacements at x, y, z, 

_ 1 d^ 1 cP<p 

~ 4 7r (A + n) dx 4:in<i dz . dx^ 
1 r/x 1 d^ 



4 TT (A + n) dy 4 irN dz . dxf 

1 , ^x 1 ^<^ , \ + 2n 



.v>. 



47r (A + n) dz 4 7rN " dz^ ' 47rN (A + n) 
From \i, V, and iv all the strains and stresses may be calculated at 
any point. 

* The interested student may refer to a paper read by Prof. Carus Wilson 
before the Physical Society of London (June, 1891) on " The Influence of 
Surface Loading on the Flexure of Beams," and to another by Dr. Chi'ee. 



381 



CHAPTER XVI. 




A a 



-!-^-H 



6 </ r 



Fig. 206. 



BENDING. 

316 In Fig. 206, c d is a beam carrying a weight w. We 
know that the beam transmits the weight to the walls, and that 
in doing so the beam is kept in a strained condition ; we must 
consider what is the state of strain in the beam. To observ^e 
this it will be well to take a beam which is very visibly strained, 
a beam of indiarubber. a B is its appearancje when lying on 
the table ; draw upon 
it a number of parallel 
lines in chalk or pencil, 
a h, G d, e fy etc. Now 
if we support the beam 
at its two ends, and load 
it, we find that the lines 
a b, G d, etc., remain 
straight, but they are no 
longer parallel. We find 
the distance a c to be 

less than a c, but h' d! is greater than h d. In fact, a' c' is 
compressed, h' d' is extended. We find also that along the 
line e' f' there is neither compression nor extension, e' f' 
remains of its old length, although it is no longer straight. 
If we consider each cross section 
of such a beam we see that the 
upper part of it is in compression, 
the lower part of it is in exten- 
sion, and there is a straight line 
in the middle where there is 
neither compression nor extension. 
This line is called the neutral axis 
of the cross section, and all these 

axes lie in a surface called the neutral surface of which e' f' is 
an edge view. Fig. 207 is a magnified drawing of the small 
portion of the beam between two cross sections, c e f d 
shows its original shape, c e f d' its shape when strained. 
Evidently there is more compression at c 6 than at I' m. 




382 



APPLIED MECHANICS. 



The compression becomes less and less as we come nearer h' j', 
then the extension begins, and becomes greater and greater as 
we get farther awav from h' j' until we get to d'f, where it is 
greatest. If the material is likely to break in compression 
it will be most likely to break at c' e. If it is likely to break 
in tension it will be most likely to break at d'f. 

317. If we know the compression or extension at any place, 
we can calculate what it is at any other place, for the strain is 
evidently proportional to the distance f rom tlie middle. Thus if 
at e there is a compressive strain of '002, that is, there is a 
compression of '002 foot for every foot in length, then at m\ 
half-way between j' and e', there is only a strain of '001 . There 
is the same strain at i' the same distance below j', but it is now 
an extension. The material resists being strained in this way, 
and the pushing and pulling forces which it exerts at the section 
e f\ Fig. 207, are just the forces required to balance all the 
other forces acting on the part e d t/"'. 

As e T>Tf' is a body kept at rest by forces, and which is no 







w, 


w» 


Wj 




E 




f \ 




D 


-^ 


H 


r 


J 


- 

















^T 














F 








J 


S^ 


' 








P 



Fig. 208. 



longer altering in shape, it is to be regarded as a rigid body.'^ 
ISTow what is the condition under which it is kept at rest 1 

318. The beams used by us are almost never deformed so 
much as the beam shown in Fig. 206, and indeed our theories 
are only true on the assumption of exceedingly small changes of 
shape. Let, then, the part e' D 'if be drawn less deformed as 
E D T F in Fig. 208, and consider its equilibrium. ^Ye had 
better consider more weights than one, loading it. 

The forces W;^, Wn, and Wg represent loads, and P is the 

* In books on mechanics you may have read much about rigid bodies and 
the laws of their equilibrium, and yoii may have thouglit that such bodies 
had no existence ; but you must remember that we can regard a quantity of 
water, or a piece of steel spring, or a rope, as a rigid body for the time being, 
if it is being acted on by forces, and is tw longer changing its shape. 




APPLIED MECHANICS. 383 

supporting force at the end T of the beam. In Art. 99 we saw 
that if all the loads on a structure are given, the supporting 
forces may be calculated. 

319. We are now considering only horizontal beams on 
which the loads are only vertical and the supporting forces 
vertical. Students had better work again here 
a few exercises, to find the supporting forces 
when the loads are given. Either neglect the 
weight of E D T F itself or imagine it represented 
by Wo. Suppose, then, p to be known. 

The molecular forces acting on the surface 
E F (by the material to the left upon the 
material to the right of e f) balance all the 
external forces acting upon e d t f, namely, Wj, Wg, Wg, 
and p. Art. 98 gave the following three as the conditions 
of equilibrium. 

I. The upward tangential resultant of the molecular forces, 
which I shall call s, the shearing force at the section, is equal 
to Wj + Wg + Wg — p. The student ought now to work a number 
of exercises. Given loads, find supporting forces ; find s at 
many sections ; show all answers in one diagram and call it the 
diagram of shearing force. Observe that we assert nothing as 
to how this shearing force is distributed over the section. We 

shall find in Art. 369 that it is most intensely dis- 

tributed about the middle parts near the neutral "^^^ 
axis 0. ^ 

II. As the loads are all vertical there is no 
horizontal component in the resultant of the mole- 
cular forces ; in fact, all the pushing forces on 
E F balance all the pulling forces. Figs. 209, 
210 show E F magnified, its actual shape and side 
elevation also. At H, a point in e f at the 
distance y from o o the neutral axis, the pushing 
force per square inch or the compressive stress ,j^ 
being proportional to y, let us call it c y, when ^_. 
c is some constant ; if a point is at h' (Fig. 209), ^j^ 210 
we shall call h' a negative value of y, so that a 

pulling force is regarded as the negative of a pushing force. Now 
at H let there be an exceedingly small portion of area a, the 
force on this is c y a, and we must have the sum of all such 
terms as c 2/ a for the whole area to be zero. This is only another 
way of saying that the pushing and pulling forces are equal. 



384 



APPLIED MECHANICS. 



All the terms c y a have the same multiplier c ; hence wliat 
we state is that if every little portion of area a be multi- 
plied by its y, the sum is zero. When this is so, Art. 109 tells 
us that 0, the neutral axis, must pass through the centre of 
gravity of the area. This is why we are always so anxious to 
find the centre of gravity of the section of a beam. The rules 
for finding the centre of gravity of the area are given in 
Art. 111. We are now about to find the value of c. Notice 
that if we know c we know the stress at any point in the 
section, and we particularly want to know c. o E or c. o F, the 
greatest stress. 

III. The moments of all the molecular forces about any axis 
balance the moments of Wj, Wg, Wg and p about the same axis. 
Now as these are all vertical forces, if we choose an axis at 
right angles to the plane of bending (the plane of the paper) in 
the plane e f, notice that about any such axis these moments 
will be the same ; hence we speak of the moments of w^, Wg, Wg 
and p about any such axis as the bending moment m about the 
section e f. When it tends to make the beam convex upwards 
I call it positive. Hence M=: -P'TF +Wj-ie + W2'Je4-W3-le 

The student ought to practise the calculation of m for many 
sections of a beam and then shov*^ his answers in a diagram, 



^ 



^^ 



^ 




Fig. 211. 



We have an easy graphical method of drawing such a 
diagram {see Art. 349). We call it a diagram of bending 
moment. 

320. Very well, then, the sum of the moments of all such 
molecular forces as c y a must be equal to m. Take the moments 
about o, the neutral axis; the moment of cyai^cya x y 
or c y^ a. Now, when every little portion a of an area is multi- 
plied by the square of its distance from a line and the sum 
taken, it is called the moment of Inertia I. of the whole area 
about the line, and it is easy to find I. for any area about the axis 
O 0. Part of Chap. YII. is devoted to this subject of moments 
of inertia. We have, then, c I — ai, oi* c = m/I ; and hence the 
compressive stress p at points y inches from the neutral axis is 



APPLIED MECHANICS. 



385 



;? = M 2//I . . . (1). As ?/ is greatest at points like e or F, we have 
the greatest compressive and tensile stresses at these points. 
In fact, the greatest compressive stress in the section is at e, 
and its amount is o e. m/I . . . (2) ; the greatest tensile stress is 
at F, and its amount is o f. m/I . . . (3), and these two expressions 
give us the great laws of strength of beams. If we know the 
stresses which the material will stand, we know whether the 
section e f will withstand the bending moment M. 

321 . If the material is cast iron it is advisable to have o e = 
about 4|- times o f, because cast iron will stand about 4|- times 
as much compressive as tensile stress. Hence 
the usual economical cast-iron sections are as 
shown in Fig. 211, with centres of gravity 
near the bottom boundaries of the sections. 
Whereas in wrought iron and mild steel and 



other materials the resistance to compressive 

stress is much the same as the resistance to 

tensile stress, and consequently o e is made 

equal to O f, and the 

usual economical sections 

are as shown in Fig. 221. 

322. In the model, Fig. 
212, which shows a beam 
fixed at one end and loaded 
at the other, part of the 
aiaterial has been removed, 
ind instead of it we have 
inserted a chain or link a, 
which is only capable of 
which is only capable of 




Fig. 212, 



exerting a 



pulL 
push. 



and 
It 



horizontally on m N o F 



a rod B, 

is found that forces acting merely 
are not sufficient to keep it at rest ; 
we also need an upward force at M F, which is equal to the 
weight w, together with the weight of M N o f itself. We see then 
that at such a section as M f of a beam we need pulling and 
pushing forces, but also to satisfy the first condition given above 
we need the shearing force at m f. In fact, an upward force w' 
must be exerted at m f equal to the weight w and also to the 
weight of M N o P. At M F the bending moment is w x o f, 
together with the weight of m N" o F x the distance of its centre 
of gravity from m f. This is to be balanced by the pull in the 
chain a or the push in the rod b, for these are equal, multiplied 



OOO APPLIED MECHANICS. 

by the distance between thek lines of action. If a beam is 
long, the shearing force exerted by the material at a section of 
the beam is usually not so important to consider as the pushing 
and pulling forces, and in many cases it is neglected. When a 
beam is very short the shearing force becomes more important 
to consider. 

323. We shall now take a casein which there is only bending 
moment to be balanced by the material at a section. Let A b 
(Fig. 213) be a strip of wood or metal originally straight, whose 
weight we shall neglect. Fix or solder to the ends stout pieces 
of metal, and by means of cords and weights, or in any other 





Fig. 213. 



Fig. 214, 



way, exert couples on these ends as shown. Consider now the 
equilibrium of any portion say, c D B (Fig. 214). At the sec- 
tion C D we know that pulling and pushing forces must be 
exerted by the material which exists at the left of c d on the 
material which exists at the right of c D, and the moments of 
these just balance the moment of the forces f and f, and this is 
evidently the same at any section of the strip. The bending 
moment at any section is then the moment of the couple or 
torque M acting at either end. Magnifying the section e f, as 
in Fig. 210, and representing the amounts of the pulling and 
pushing stresses by arrows, we see as before that as the sum 
of all the forces one way must be equal to the sum of all the 
forces acting the other way, and as the stress at each place is 
proportional to distance from o, the part where there is no 
stress or the neutral axis is as before, a line through o 
at right angles to the paper, and this must pass through 
the centre of gravity of the section. We see also that the 
stresses at all points of the section are given by (1), and if Ave 
particularly desire to know the greatest stresses they are 
given us by (2) and (3). 

324. Unsymmetrical Bending.— If the tending moment m in a 
section is ahout an axis o x^ through o the centre of gravity of the 
section, and if this is not a principal axis [see Art. 114), hut makes 



APPLIED MECHANICS. 



387 



the angle x^ o x = a with a x^rincipal axis o x (about which the 
moment of inertia is ij), the other being oy (about which the 
moment of inertia is I2), then m may be resolved into m cos. a about 
o X, and ai sin. a about o y. The stresses and strains due to these 
sej)arately have now to be combined y 

to obtain the actual stresses and strains 
due to M. Thus at a point whose co- 
ordinates referred to the principal axes 
are x and y, we have the total stress 



y M sm. 



f 



If / is put equal to 0, we find the 
position of the neutral line. It evi- 
dently is like o a, and makes the angle 
j8 = G o X with o X, such that tan. 

j8 = -^ tan. a. It is now easy to find 

the points which are at the greatest 
distance from o q, and these are the 
points at which there is greatest stress. 




Fig. 215. 



Exercise. — In a beam of rectangular section, if the axis about which 
the bending moment takes place is at right angles to a diagonal, show 
that the greatest stress is at corners, and is of the amoimt 6 m -^ ^(?d if 5, 
d, and d are the breadth, depth, and diagonal of the rectangle, 

325. The line which passes through the centre of gravity 
of every cross section, being neither extended nor compressed, 
is of the original length of the strip. When the beam is bent 
as in the figure, A B becomes longer than this, and a b shorter, 
yet their ends are in the same planes A a and B h. Thus the 
strip may be considered as a bundle of fibres lying in arcs of 
circles which have the same centre and subtend the same angle 
at that centre. If we know their relative lengths we can tell 
where the centre of the circle is, Now we know the stress per 
square inch on a certain fibre, and we know its original length, 
hence we can calculate its present length (see Arts. 241 and 265), 
and its length is to the length of the neutral fibre as its radius 
is to that of the neutral fibre. In this way we find that the 
radius of the neutral fibre is numerically equal to the modulus 
of elasticity of the material rmdtijjlied by the moment of inertia 
of the cross section, and divided by the bending moment at the 

1 M 

section, or the curvature - = —••• (4). 

To put it in another way : — The curvature in Fig. 207 being 



388 APPLIED MECHANICS. 

angular change per unit length, is the angle between c' d! and 
e f if h' j' is unity. But this angle is, strain at y -^ y, or 
/:» -^ E2/ . . . (5) and by (1) this becomes (4). The form (5) is 
sometimes useful in indicating the greatest curvature which 
may be given to a beam without hurt, by making ^; the greatest 
stress which the material will stand, ?/ being the greatest distance 
of any point in the section from the neutral axis on the com- 
pression or tension side. In the same way it is easy to see that 

if a beam was already bent, having a curvature — , the change 

of curvature — - -^ = — ... (6). 
r rg E I ^ ^ 

Example. — A straight strip of tempered steely 0*7 inch 
broad, 0-1 inch thick (this represents the deiHh of a beam), is 
subjected to a bending moment of 100 pound inches : find its 
radius of curvature. Answer : — The moment of inertia of the 
section is 0-7 x -1 x -1 x -1 4- 12, or -0000583. The modulus 
of elasticity of steel is, say 36,000,000, and 36,000,000 x 
•0000583 -^- 100 gives 21 inches for the radius of curvature of 

the bent strip. The curvature is ^j. If the strip was already 

bent before the load was applied and had a radius of curvature 

50 inches, then the chansfe of curvature - ~ ^77— ?ri j hence 
' * r oO 21 ^ 

r =z 14*8, or if the new bending takes place in an opposite way 
to the old so that the new curvature is called negative for our 

purpose, -7- — - =jrr, so that r = — 36-2 inches. 
50 r 21 

JExercise. — A beam is supported horizontally on two points, one under 
each end ; c is a point of the beam one-fourth of its length from one of 
the points of support. Compare the curvature at c, supposing thcbeam 
to be uniformly loaded, with what it would be if the beam were without 
weight and the load concentrated at the middle point, the total load in 
both cases being the same. Ans., As 3 to 4. 

326. When a beam is loaded in any v/ay, we know now how 
to find the bending moment at any place, and if we know the 
modulus of elasticity of the material, and the moment of inertia 
of the section, we can find the curvature of the beam. We 
may draw a bent beam, then, in the same way as we draw the 
springs of Fig. 216, but the beam is so little curved usually that 



APPLIED MECHANICS. 



389 



we have difficulty in getting compasses long enough. In this case 
it is usual to diminish all the radii in some large proportion, 
remembering that the deflection of the beam as you draw it is 
increased in this proportion. For a beam fixed at one end and 
loaded at the other we get a curve just like the portion s t in 
Fig. 216 c, s being the fi.xed end and t the loaded end. 



The following method is not perhaps so instructive for 
beginners, hut it is the quicker and more accurate method. The 






590 APPLIED MECHANICS. 

theory is given in Art. 358. Suppose a c d z b is a diagram showing 
the value of m -r- e i at every place, e being Young's modulus, and 
I the moment of inertia at each section. Treat it as if it were a 
diagram of loading, as described in Art. 350, and proceed as if you 
wanted to find m ; in truth what you will find is a diagram which 
shows everywhere the deflection of the beam. That is, if you get 
fl ^ of Fig. 349 horizontal, you will have foimd the shape of the 
beam turned upside down. The scale will be : — If the beam is 

drawn to a scale of 1 inch represented bv n inches, and if ^- is 

^ * ' E I 

drawn to a scale of imity in poimds and inches represented by 
y inches, and if an area of 1 square inch on the diagram of - — 

^ ° EI 

in Fig. 237 is represented on the diagram of Fig. 236 by v incbes, 
then a deflection of the beam of 1 inch is represented by a distance 
of y n^ v/o H inches, o h is to be in inches. 

327. Elastic Curve. — If we take a straight uniform strip of 
steel and subject it to two equal and opposite forces in the same 
straight line, the strip will assume one of the forms shown in 
Fig. 216, which all go under a common name — the elastic curve. 
Now consider the part p b, Tig. 216 a. Neglecting its own 
weight, it is acted on by a force f at b, and at p there must be 
an equal and opposite force to produce balance. There is a 
force at P tending to compress the steel, but what is of more 
importance is the fact that f produces a bending moment at p, 
and the amount of it is the force x the distance p k. Now our 
strip is everywhere of the same material and section^ and the 
only thing that can alter is its curvature. This curvature at 
any place we know to be greater when the bending moment is 
greater, and less when the bending moment is less ; in fact, the 
radius of curvature is inversely proportional to the bending 
moment, and this really comes to the fact that the radius of 
curvature at any place p is inversely proportional to the 
distance P K ; or if the distance P k of any point from the line 
of force is called x, as r = e i/m or e i/F.a.*, and as e i and f are 
constant, r x is constant. 

We can obtain the shapes shown in Fig. 216 in two ways : 
first, by taking a straight strip of steel and performing the 
operation ; secondly, by drawing the curves in a series of arcs 
of circles. Suppose we have calculated, as in the above example, 
that the modulus of elasticity of the material multiplied by the 
moment of inertia of its cross-section is, say, 200 in inches and 
lbs., and suppose we know that the force acting at b is 10 lbs., 
then we know that the radius of cui-vature at p is equaJ to 200 



APPLIED MECHANICS. 



391 




Fig. 217. 



divided by the bending moment at p, which is 10 x p k. In fact, 
the radius of curvature at p is equal to 20 divided by p K or x. 
Choose now in Fig. 217 the point c as the middle point in the 
strip. Suppose c d or the vahie of x for c to be 4 inches, then 
the radius here is 5 inches. Take c o = 5 inches, and with o 
as centre describe a small arc, C E. Join 
E and produce it. Now at e measure 
E F, the value of x for e, and suppose you 
find it 34 inches; divide 20 by 3-4, and 

we get 5*88 inches, and set this new radius .d., 

off from E to o'. Take o' as a new centre, 
and describe the short arc e G of any con- 
venient small length, and in this way proceed 
until the curve is finished. This is not a very 
accurate method of drawing the curve unless 
the arcs are very short, and small errors are apt to have 
magnified evil effects, but I know of no better exercise to 
impress upon you the connection between radius of curvature 
of a strip and the bending moment which produces it. You are 
therefore supposed to have actually drawn one such curve at 
least before proceeding with your study of this 
subject. There is a way of diminishing errors, easy 
to discover for yourself if you are interested. 

328. Parts of these curves happen to be the shape 
taken by liquids, because of their capillary action, 
between two parallel solid faces. They are also the 
shapes of the arches which are best fitted to with- 
stand fluid pressure. Thus, for instance, in Fig. 218 
the curve from m to n is of the shape of the curve 
Fig. 216 e, from m to N, the free water level being the line A B ; 
and in Fig. 219 the middle line of the joints of the arch m to N 
is the same curve inverted. The water, whose pressure it 
resists, has as free water level the line A B in the relative posi- 
tion of the line of force to the elastic curve. 

329. When a strip of elastic material is bent, it not only 
alters its shape in the well-known way, but it alters the form of 
its cross-section. On the convex side of the strip the breadth 
becomes concave, and on the concave side of tlie strip the 
breadth becomes convex. It is very easy to try this for your- 
self on a broad strip of steel or a bar of india-rubber. These 
saddle shapes of the surfaces are due to the fact that when each 
fibre is pulled it gets thinner as well as longer {see Art. 265), 




21S. 



392 



APPLIED MECHA>?1CS. 




and when it is pushed it gets broader as well as shorter, and it 
is very curious that this action should not interfere perceptibly 

_^_^__ with the laws of bending as I 

^^^^^^^j ; ^ Sf^^^^^P have given them to you. In 
=0:4g^^^^^Z^-^^i^^^^^^ all probability it does so in- 
terfere, however, in the case 
of a very broad thin strip of 
material greatly curved. 

330. In any section of a 
beam, con.<ider a rectanguLir 
portion of area of breadth h 
(parallel to the neutral line) 
and depth, d. It evidently 
undergoes a change of shape of outline whose character does not 
depend upon what portion of the section it is in. But, for 
ease of calculation, let us take it to be symmetrical above and 
below the neutral line oo (Fig. 220). The 
curvature of the beam is m/ei. The com- 
pressive stress at a distance y from the neutral 
line is / = m y i. This produces compressive 
strain, but it also produces a lateral thickening 
strain j8/ {see Art. 265). Hence the breadth 
b of the rectangle at any place y becomes 
broadened by the amount b m ?/)3/i, and hence 
the straight sides a c and b d alter to the 
straight sides e h and g l. Also the dimensions 
parallel to a o and b o on the a b side of o o 
lengthen equally, and on the c d side they 
shorten equally. Consequently a b and c d be- 
come arcs of circles with the same centre, 
y is J<f at AB, and - |<? at cd. Consequently 
EG = J (1 + M i<7/3/i), HL = * (1 - -SLld^li). And if r is the 




Fis. 220. 



radius of o' n o 



^-V, and hence - 
1 - ^M^iS/i d 



■ r - ^d 1 - iMdfi/f d SLdp; 

The student will recollect that we used o in Art. 265 to 



represent — ; and j3 is the smaller number, such that ;3/a was called 

Poisson's ratio a. We see now that the curvature of o' n o' is <t 

times the curvature of the neutral line of the beam. In the above 

case the curvature of either the top or bottom surface, or of the 

neutral surface, is anticlastic or saddle-shaped. 

Exercise. — One side of a plate of metal is at 0^'c and the 

other is at 0o°c. The plate when cold is plane. What is now 

its curvature, and what is the greatest stress in the material 

if curvature be prevented? Ans. — a^. (0^ — 0o) ^jy^. (This 

measures the specific curvature of the surface, being the 

product of the two principal radii of curvature at any point 

of the surface.) Greatest stress =Ea {dc, — Q^. {See Art. 5.) 



393 



CHAPTER XVII. 

STRENGTH AT ANY SECTION OF A BEAM. 

331. Observe that in any section of a beam, the stress being 
greatest at places furthest away from o o (Fig. 220), these are 
places where the material is most useful in resisting bending. In 
some cases, as in railway girders, economy of material and its 
weight are so important that there is very little area of section ex- 
cept at E andF, where there are tWO booms or flanges each of area 





mm^ «=^ 



( m^mk a 




Fig. 221. 



A : one where nearly all the compressive forces act (/a is the total 
compressive force if /is the compressive stress), and one where 
nearly all the tensile forces act (being equal to the same /a), 
and the sum of their moments about o (or any parallel line at 
right angles to the plane of bending), being /Ad ii d is the 
distance from e, the centre of one boom, to f, the centre of the 
other (and called the depth of the beam), is equal to the bending 
moment. In plate girders there is a thin web, or perhaps two 
(Fig. 221), holding the booms or flanges in their proper posi- 
tions ; in open-work girders there are diagonal braces to do 
this. The function of the web or diagonal braces is to resist 
the shearing forces, and we have good reason to know (see 
Art. 334) that the booms or flanges need be proportioned only 
to withstand the bending moment. In Fig. 100 we had a 



394 



APPLIED MECHANICS. 



IfTTTTTIXI 



f^ 



girder with few diagonal pieces. Noav we already know that 
for absolute certainty in calculating the forces exerted by the 
pieces it was necessary to imagine pin joints not merely between 
the braces and the booms, but between one piece of boom and 
the next. Evidently this is nothing more than assuming that 
each piece can only exert a pulling 
or pushing force, and has no shear 
in a cross section. It will be found 
in Art. 369 that there is practically 
no shear in the cross section of the 
flanges of even a plate girder, and 
that the above easy practical rule 
(/ac? = m) used so generally by en- 
gineers is quite legitimate. 

332. When the web is a plate it 
is hardly worth while to calculate 
whether it will resist the shearing 
force ; we always find that the web 
which is thick enough (never less 
than I inch) to let the girder be 
handled, and to let other girders be 
fastened to it and pieces to prevent 
buckling, is ever so much larger to 
resist shearing than is merely ~ ne- 
cessary for this purpose. Plate 
girders are used up to 75 and even 100-feet span for railway 
bridges. For great spans, as in the Menai tube of Fig. 222, 
this form is no longer thought to be economical. 

In open-work girders it is 
necessary to make the calcula- 
tion, and it is done in the 
following way. In built-up 
structures, if iron and timber 
are used, it is well to use 
timber for the struts (see 
Art. 372), because it is late- 
rally large, and iron for the 
ties. In structures in which 
different materials are used, expansion is different in the 
different parts, and there ought to be no redundant parts. 
Thus if four pieces form a parallelogram and it has two 
diagonal pieces, of the six one is redundant. If all expand 



I T T J, M I J 



222. 



1^1^ 



Fig. 223. 



APPLIED MECHANICS. 395 

equally there is no great harm, but if one expands more than 
another great weakness may result. 

Exercise. — A hinged square of four pieces of iron, each 6 feet long, has 
two diagonal pieces, one of brass and the other of iron. The iron bars 
are of equal cross section, and three times tLat of the brass one. There is 
no initial strain in them at 0° C. What are the forces in the pieces 
at 30° 0. ? 

333. Redundant parts are often useful in stiffening a struc- 
ture when the loads are apt to be very different at different times. 
When it is necessary to reject redundant parts we need no 
rules ; common-sense Avill tell us, for example, that if the bars 
are long and slender we ought to reject the struts rather than 
the ties. When there are no redundant bars, the student 
follows the graphical method of Chap.YIII., or the correspond- 
ing analytical method which gives the same answer. When there 
are redundant bars he 
must make certain as- I, 

sumptions. Thus, if a ^ ^^>|B 

table with fairly uni- 
formly distributed load 
upon it has 20 legs, we 
usually assume that each 
leg has a twentieth of 
the load if they are all 
equal in length and the 
floor equally yielding 
everywhere. Probably Fig. 224. 

this is wrong, and if we 

had sufficient information we might see reasons from theory oi 
common-sense to suspect more load on some of the legs than on 
others. But we can do no better. If in Fig. 224 b c and d e 
are the booms in compression and tension, if b d is h inches, the 
vertical distance between the centres of area, and if the sectional 
area of each is A, then, as we have seen, /a A = the bending 
moment at the section b d. We are supposed to know s, the 
total shearing force at the section. This is the force with 
which the material to the left of b d acts upwards on the 
material to the right of b d. If the push or pull in a diagonal 
piece is p, the upward component of this is p sin. 8 if is its 
inclination to the horizon, and the total shearing force is resisted 
by the vertical components in the diagonal pieces. Thus if 
aU the diagonal bars are of iron, and if they are equally inclined 




396 APPLIED MECHANICS. 

at 45° to the vertical, if the force in each of them is p, the 
upward component of each is P cos. 4-5° or p/y.T ; if there are n of 
them crossing the vertical plane B D then n p/v/'q is the whole 
shearing force s. If s is known, p may be calculated. When a 
figure is drawn it is easy to see which piece exerts a push and 
which exerts a pull. 

Exercise. — In Fig. 224 there are four diagonal pieces crossing 
the section at 45°. If the bending moment at b d is 3 x 10^ 
pound-inches and b D is 5 feet, find the area of the boom at b or 
at D. If the shearing force at b d is 4 x 10^ pounds find the 
probable push or pull in each of the four diagonal pieces and 
distinguish struts from ties. Ans'.^ 5 x 10^-^/; 7*07 x 10*. 

If instead of Fig. 224 we have a section of a Warren girder 
with only one diagonal piece crossing the section at an angle of 
60° with tlie horizontal, w^hat is the push or pull in iti Ans.^ 
4-62 X 10-5. 

Exercise. — If only one diagonal piece at 45° meets a boom 
and a vertical piece, show that the force in the vertical piece is 
equal to the total shearing force in the girder section at the 
place. 

As the student must by this time have drawn diagrams of 
shearing forces, he knows that when the load of a beam is 
uniformly distributed over it the shearing force is greatest at 
the left end, diminishes to zero at the middle, and increases 
positively to the right-hand end of the beam. The diagonal 
pieces are therefore large at the ends of abeam and small in the 
middle, and especially in large girders, for much of the load of 
any large girder must be distributed uniformly. 

334. We see that when, as in large girders, we think greatly 
of economy and we know our loads and that they are vertical, we 
have flanges or booms whose sizes may be calculated with a fair 
amount of correctness. Even in smaller beams to carry vertical 
loads, it is convenient to look at what occurs at a section from 
this point of view ; that the flanges resist the bending moment 
and the web the shearing stress. Taking, in Fig. 223 for 
example, h the vertical distance in inches from the middle of the 
top flange to the middle of the bottom ; A,, the area of the top 
flange, Aj. the area of the bottom one, in wrought iron we 
make Ac = At;, because y^ is much the same asy^, but in cast 
iron/c is 4|- times /t, and hence we make 4 J a^^ a^; or the 
bottom flange, which is in tension, 4 J times the area of 



APPLIED MECHANICS. 



397 



The bending moment is /. a^. or /^ a^ multi- 



the top one. 
plied by h. 

Exercise. — The top and bottom flanges of a rolled section 
of wrought iron are 8" X f ". The web is of same thickness. 
The height over all is 12". What is the bending moment when 
the greatest tensile stress is 10,000 lbs. per square inch *? 

Work this in two ways. I. The tensile or compressive force 
in each flange is 8 x | x 10,000 = 50,000 lbs. The value of h is 
12" - ^' or llf. Hence the bending moment is llf X 50,000, 
or 569,000 pound-inches. II. The moment of inertia of the 



section about its neutral axis is ^^J±-mi 

i^ 12 



388. 



T 



This divided by 6 is z, the strength modulus, and 10,000 z, or 
647,000, is the bending moment. 

335. The above example gives a fairly good idea of the error 
in adapting the usual practical rule for a railway girder where 
there is almost no web to a rolled girder where there 
is a web. The web is here distinctly 
of importance in resisting bending 
moment. Some engineers, instead 
of taking h from centre to centre of 
flange, take it the height over all ; 
but even if we take A =12" in the 
above we still have an answer , 
which is about 7 per cent, too small. 
The error is on the safe side. 

Nevertheless, it is always better 
to keep to the correct rule of 
Art. 320 in girders which have an 
important web, and in all mechanical 

engineering calculations we keep to the correct rule. I 
therefore give here a list of the values of i and of z for various 
forms of section. 

Exercise. — Show that the centre of gravity o of the area in 
Fig. 225 is 2 inches above the bottom. Take it as x inches. 
The middle of the bottom rectangle is cc ~ f from o and the 
middle of the top one is 3 J - a; from o. Hence {x—\) 5 = 
(3J — £c) 5 or x^2 inches. The moment of inertia of the top 

5 X 1^ 

rectangle about axis o o is -r-^ + 5 {l^Y "^ 2 If, The moment 

12 



s" ^ 

Fig. 225. 



398 



APPLIED MECHANICS. 



5 X P 

of inertia of the bottom rectangle about axis o o is — -j- + 5 

( ] |)~ == 1 If : and the sum is i = 33i. z^ for the top is i -h ^ = 8 1 
Zo'for the bottom is i -4- 2 = 16f. 

TABLE YI. 



Moment of Inertia of Section. 



-i-..> 






12 




12 



"I 
f 



^^ 






strength Modulus of Section, 






i*^ 



A ' 2 s3 
12 



6u 






APPLIED MECHANICS. 



399 



Moment of Inertia of Section. 




(b h^ - bh^-f - 4 B H ^A (h - hf 
12 (bh - hh) 





' 36 {b + b{) 




[ ^h (^3 - h,^) + b^ {^hi + x^-^^i 







z = ijy. 
Strength Modulus of Section. 



(rh^ - hh'^f - 4EH&A (h - hf 
6 fBH2 + ^7^2 _ 2JhA) 



24 
12 



12(2i + ^'i) 
■ 12 (J + 2 bi) 



X 



400 



APPLIED MECHANICS. 



Moment of Inertia of Section. 



z = i!y 

Strength Modulus of Section. 



[ 



l\b [r^ - Ai3) + b^ (A^a + h.^) + ^2 



(-^-fli^ 



:...^- 






ux^a 



4 __,^„ 64 



D4 



^^-l^i-^-^ 





— (r,D3 - b^) 



(Paracolic segment) 



t! = 



^'=1 



z' = 



32°" 



32 ■ D 



32 



VA D / 






APPLIED MECHANICS. 401 

Exercise. — Find tlie greatest load that may he imiformly distributed 
on a cast-iron girder having top and hottom flanges tmited by a web, of 
the following dimensions :-r- Width of upper flange, 3 inches; of lower 
flange, 9 inches; total depth, 12 inches; thickness of each flange and of 
the web berag 1 inch; distance between the points of supports, 10 feet; 
when the greatest admissible stress in the compression flange is 6 tons per 
6i"iare inch, and that in the tension flange IJ tons per square inch. 

Ans., 9-8 tons. 
336. Proportion of Depth to Length in Railway Girders. — It is 
usually assumed that maximum economy of weight in booms and 
diagonal pieces leads to a most economical ratio of depth d to 
length I, but we must confess that we feel dissatisfied with the easy 
mathematical statements sometimes deduced on this subject from 
incomplete data. Take it that in girders of the same style the 
diagonal pieces make some known angle with the horizontal. Let 
us take 45°, for example. Then each bar of length d \/ 2 and 
cross-section a withstands a shearing force s = a/" x V 2, where / 
is the shear stress, and has a weight d^/ 2 . a ('28), or the weight 

is dV2 ('28) - . The weight of the corresponding pieces of 

boom is 2^ A (•28), where Afd = m, where / is also the tensile 
stress. Total weight of a bay is therefore 

2^(-28)^ + 0^-)2^s....(l), ..f- 

or the weight in pounds per inch run of the girder is 

f(^s)....,, 

We see, therefore, that if the inclination of the diagonal pieces is 
fixed, the gTcater d is the better ; and there is no most economical 
depth of a girder derivable — at all events, from these simple 
considerations. It is true, however, that making the depth great, 
if the inclination is constant, means that we are increasing the 
length of the unsupported part of the compression boom and struts, 
and they may need more lateral bracing ; or, as it is better to put 
it, the cost of the compression members per pound will increase. 
Again, the weight of the platform mil also increase. These are, 
however, questions of a different kind, difficiilt to settle by 
elementary mathematical equations when systems are so different. 
Writing down such general expressions as we may, there is 
e^adence that there is greater economy in weight, whatever there 
may or may not be in cost, in letting the depth get less where the 
bending moment is less, instead of keeping it constant. 

The most important matter, how the natural period of ^dbration 
of a bridge ought to come in, seems never to be brought forward in 
these calculations. Professor Milne finds that the horizontal trans- 
verse deflection is the most serious motion of a railway bridge. It 
begins when the train is perhaps 200 feet or more away ; it 
becomes accentuated with every passing carriage, and when the 



402 APPLIED MECHANICS. 

whole train has passed there is the natural vibi-ation of the hridge, 
which continues for some time. These vibrations are due to lurch- 
ing of carriages and impact of wheel-flanges against the rails. 
Sometimes a Hght waggon seems to produce much larger effects 
than the heavy loc motive, and there is some speed of train with 
which the vibration is much more serious than with quicker or 
slower speeds. Bridges have not yet been studied from this point 
of view, and engineers must for the present rely upon their large 
factors of safety. "We are sure that more attention ought to be 
paid to these lateral vibrations, which seem to be greatly accen- 
tuated when gusts of wind are acting laterally. 

The Board of Trade rule is 5 tons per square inch on wrought 
iron and 6J on steel. This is not sufficiently safe. It is well to 
say in tension on iron, 4 in compression, and steel, say, | greater. 
In some large bridges it is estimated that stresses due to wind are 
greater than those due to rolling loads. 

The force acting on the rails in the direction of their length is 
sometimes as much as J^th of the weight of a train if the train is 
quickly stopped. 

As for the vertical motion, its consideration w-ill probably lead 
to some rule connecting maximum deflection w^ imder static load, 
and length of girder I. There is a vague sort of imderstanding that 
Vi shall be something between ij^^^th and x eVo^ii of ^- ^ beams are 
of imiform streng-th and depth, d, the curvature is constant, being 
'2fJT.d {see Art. 362), / being the greatest stress in every section; 
and hence in girders supported at the ends the deflection iji is 

equal to l-f/idE. If, now, we take ^/^ to be //1, 200, ^ = ,- 

or^ = ^~y. Taking E//as 3,000, ^ = 10 (3). 

It is not quite fair to say that the calculation of the probable 
loading of railway bridges is more scientific in America than in 
Britain. British engineers differ much more in their assumed 
loads than American, but in neither case can it be said that there 
is a scientific basis for the riiles in use. It is very important that 
the student should know this, because there is sometimes a pre- 
tence of accuracy of treatment in bridge calculations which is 
quite misleading. The following rules are more common than 
others, and may well be used in academic problems. 

In ordinary railway bridges the greatest possible rolling load 
may be taken as if it were a static load of iv-^ tons per foot-run for 
a double line, where Wj = 3§ + 176//, if / is the span in feet. 
This is really on some such assumption as that a rolling load must 
be multiplied by 1 J to convert it into the equivalent static load. 
The dinunution with length is due to the fact that the engine 
weight is more intense per foot than other parts of the train. The 
weight of the platform may be taken as ic., = 0-7 -f '0072 I tons 
per foot of the span for a double line, u:^ increases with the span 
because of greater wind-bracing and the greater width of larger 
spans necessary for lateral stability. 



APPLIED MECHANICS. 403 

Half of every term may 'be taken for a single line, and used 
even as low as for 15-feet spans. A railway girder is usually 
built with so mucli negative deflection or " camber," as it is called, 
that it will become just about level when loaded. 

From (2) of Art. 336, taking m as proportional to tvl^, if to is 
the total load per foot run, and s as proportional to ivl, the whole 

weight, w.^, of the girders per foot run is equal to ( a — 4- bw.\ I, 

where a and b are two constants. If we take it, as is usual, that 
l/d is nearly constant (say 10, as above mentioned), this becomes 
w^ = liv/e, where c is some constant. Now, 



IVi 



Hence w^ = _c_ _ or -—, {wi + iv^) .... (4), 
I 

and w =■ {'lo-i + uk,) / M Y 

Whatever may be the worth of this reasoning, it gives us a rule 
which is taken to hold in all railway girders. The value of c is 
about 1,000 feet for ^Dlate web girders, 1,200 for lattice girders of 
ordinary construction, and 1,400 feet for bow-string and well- 
designed lattice girders. 

337. Small pistons for cylinders up to 20 inches diameter are 
packed so as to be steam-tight in the following way. The method is 
quite wrong. Suppose the cylinder is to be 15 inches diameter ; there 
are two or more grooves turned in the piston block, about ^ inch 
or more broad and | inch deep, to receive cast-iron rings. Many 
such rings may be made at the same time. A hollow cylinder of 
cast iron is turned up about 151 inches outside and 14^ inches 
inside, and many rings are cut from it. Each ring has a piece cut 
out, so that when it is sprung into place in the piston groove it 
may be jammed into the cylinder, its ends coming now close 
together. The cylmder keeps it smaller than its unstrained size, 
and the pressure all round is assumed to be uniform. 

Exercise for Students. — Prove that the pressure is not uniform 
all round.* 

* To show that the pressure cannot be uniform all round in the usual methoc 
of manufacture. The ring unloaded is of a circular shape ; loaded, it is circular 
and of smaller radius. The bending moment must therefore be constant. 
Let us try what distribution of pressure p will produce a constant bending 
moment ; that is, using the symbols which follow above, if Mo is the bending 

re 

moment at the end where 9 = 0, Mo + r^ 1 p sin. (p . d^ =: M, a constant. 



./ 



Differentiating with regard to 0, we find p sin. =: ; and as this is true for 
ill values of 0. we must have p = everywhere. That is, the shape can only 
je maintained by couples applied at the two ends. It is quite a usual thing 
to see workmen beating such a ring out of shape near the joint, after it has 
been m.anufactured, because it is known that, so far from pressing uniformly 
all round, it does not even fit the cylinder, but concentrates its pressure at 
certain points. 



404 



APPLIED MECHAXICS. 



The following method of making a piston ring proauces uni- 
form pressure all round, even if the section of the ring varies very 
greatly. A ring is cast which is larger than what is required ; a 

piece is cut from it, and the two 
free ends are brought together 
by a clamp. This clamp pulls 
on both ends, and the more 
nearly the resultant pull is tan- 
gential to the mean cylindric 
surface at the end the more 
nearly perfect will the ring be. 
The ring is now turned up to 
the size of the cylinder and 
finished. It is undamped, and 
may be sprung into place. 

To prove that the pressure 
must be uniform all round. In 
Fig. 226 let bpa be part of a 
piston ring constrained to be of 
the size of the cylinder, of 
radius r. It may either be kept 
in its present shape by the equal 
and opposite forces f at its ends, 
or by pressures 2^ l^s. per inch 
of its length. Let us suppose at 
first that j5 may not be the same all roiind, being some function 
of the angle a o p = 0. Draw p q perpendicular to o a. f with- 
out 2^' produced bending moments at all the sections of the 
ring ; the pressures p without f must produce the same bending 
moments. Let a r rrr ^, where p and r are any two places on 
the ring, and a is one end. Bending moment at p due to f is 
M = F . A Q, or 




M = Fr (1 



(1). 



The pressure at r on the element r . 5^ is 2)rZ(p, and the bendint 
moment duo to this at p is^Jr- . 5^ . sin (0 — (p), and we require 



i: 



pr- sin.(0 — 0) . d-p =■ -£ r {\ - cos.0) 



(2). 



In the integral 2) is a function of <p. Now we know that in general 

x) . dXf 
J 
and hence equation (2) is 

re 

r^ \ p sin.0 . rf</) = Fr(l - cos.e) .... (3). 



Ca fa 

fix).dx^ f{a- 



i 



Differentiating with regard to 0, we have r'-7>'sin.0 = Fr sin.0, or 
^ = F/r, a constant .... (4), 



APPLIED MECHANICS. 405 

It will be observed that the pressure ]) is uniform all the way 
round, even if the ring- varies greatly in section. It is quite true 
that the proof assumes the thickness to be everywhere incon- 
siderable. If, however, we assuine a uniform thickness, it is easy 
to see that if the resultant force f exercised by the clamp acts on 
each end exactly at the mean radius, there is absolutely a constant 
pressure per inch all the way round. In Japan, twenty years ag-o, 
Professor E. H. Smith told me that the above proposition was 
true. I worked out the proof very easily. I do not think that it 
has been published before. 

The usual method of making the rings is very much more 
mischievous than it may appear to be on a hurried examination. 
It seems to me that if the correct method is adopted, care being 
taken as to the proper method of applying the clamp, piston rings 
may be made in this way for the very largest cyKnders, and it is 
evident that there must be a very great reduction in the cost of 
large pistons in consequence, 

338. Curvature. — The curvature of a circle is the reciprocal of its 
radius ; and of any curve, it is the curvature of the circle which 
best agrees with the curve. The curvature of a curve is better 
given as " the angular change (in radians) of the direction of the 
curve per unit length." Now draw a very flat curve, with very 

little slope i. Observe that the change in i or -^^ in going from a 

point p to a point q is almost exactly a change of angle 



r 



change in ^ is reaUy a change in f , the tangent of an angle ; but 

when an angle is very small, the angle, its sine and its tangent, 

are aU equal . Hence, the increase in — from p to q, di^^dded by 

the length of the curve p q, is the average curvature from p to q ; 
and as p Q is less and less, we get more and more nearly the 
curvature at p. But the curve being very flat, the length of the 

arc p Q is reaUy Ix^ and the change in -j divided by Zx, as Zx gets 

Ax 

less and less, is the rate of change of j- with regard to .-r, and the 

symbol for this is -j-^. Hence we may take ^-^ as the curvature 

of a curve at any place, when it is everywhere nearly horizontal. 
If the beam was not straight originally, and if y' was its small 

deflection from straightness at any point, then ^ was its original 

curvature. We may generalise the following work for beams 

not straight to begin with by using —5 [y-y') instead of -r^ 

everywhere. 

It is easy to show that a beam of uniform strength — that is, a 
beam in which the maximum stress / (if compressive, positive' ; ii 



406 APPLIED MECHANICS. 

tensile, negative) in every section is the same — has the same 
curvature everywhere, if its depth is constant. 

If d is the depth, the condition for constant strength is that 

'— . hd = + /. a constant. But - = e x curvature ; hence curva- 

ture = -^— ,. 
E . d 

Example, — In a "beam of constant streno-th. if // = —. then 

" • a -\- Ix 

3-0 = — (a + hx). Integratinsr. we find :--,■ ~ = c ^ ax ■\- \ bx-, 
ax- E o ^ 2y dx 

E 

and again, —..y — e -\- ex + \ ax- + i . h)^, where e and c must 
be determined by some given conditions. Thus, if the beam i^ 

fixed at the end, where a: = 0, and -f = ^ there, and also y = 

there, then c = and e = Q. 

339. In a beam originally straight, we know now that if a; is 
distance measured from any jjlace along the beam to a section, and 
if y is the deflection of the beam at the section and i is the moment 
of inertia of the section, then 

dx-^ EI • • ■ • ^^^' 

where m is the bending moment at the section, and e is Young's 
modulus for the material. 

We give to —, the sign which will make it positive u m 

is positive. If m would make a beam convex upwards and y 
is measured downwards, then (1) is correct. Again, (1) would 




Fig. 227. 



be right if m would make a beam concave upwards and y is 
measxu-ed upwards. 

Example 1. — UnLform beam of length / fixed at one end, 
headed vdth. weight w at the other. Let x be the distance of 
a section fi-om the fixed end of the beam. Then m = w (^ - a;) ; so 
that (1) becomes 

w dx- 



A.PPLIED MECHANICS. 407 

Integrating, we have, as e and i are constants, 

yf dx 2 ^ 

From this we can calculate the slope everywhere. 

To find c we must know tlie slope at some one place. Now, 
we know that there is no slope at the fixed end, and hence 

3- = 0, where x = Q\ hence c = 0. Integrating again, 

— y = \lci^-\x^ + c. 

To find c, we know that y = when a; = 0, and hence c = ; 
so that we have for the shape of the beam— that is, the equation 
giving us y, the deflection for any point of the beam— 
w 

y=~{\i^''- ^^^)....(3). 

We usually want to know y when x = I, and this value of g- 
is called d, the maximum deflection of the beam ; so that 

D = - — .... (4). 
3ei ^ ' 

Example 2. — A beam of length I loaded with w at the middle 
and supported at the 
ends. Observe that 

if half of this beam W 

in its loaded condition 
has a castino- of cement 






made round it so that 

it is rigidly held, the VY' 

other half is simply ^ 

a beam of length J I, 

fixed at one end and 

loaded at the other with | w, and, according to the last 

example, its maximum deflection is 

D = ^ ^^^ ' or — (6). 

3 EI 48 . EI ^ ' 

The student ought to make a sketch to illustrate this method of 
solving the problem. 

Example 3. — Beam fixed at one end with load w per unit 
length spread over it. The load on the part PQ,is«^;xPQorw 
[l - x). The resultant of the load acts at midway between p and 
Q, so, multiplying by J (/ - :»), we find m at p, or 

^ = lw {I - xf (6). 

2 EI d^^v 
Using this in (I), we have rj-^=l'^-2lx-\- x^. Integrating, 

we have — - — = Px - Ix'^ -^ \x^ -^ c. This gives us the slope 



408 APPLIED MECHANICS. 

everywhere. Now — — where x = 0, because the beam is fixed 
there. Hence c = 0. Again integrating, 




Fig. 229. 

and as y = where a; = 0, -c = 0, and hence the shape of the 
beam is 



^ 24ei ^ 



4/^-3 + iC^ 



(7). 



«/ is greatest at the end where x = I, so that the maximum 
deflection is 

1 w^3 



D = — — 3 1\ or D 
24 EI 



EI 



(8) 



if "w = wl^ the whole load on the beam. 

Example 4. — Beam of length I loaded uniformly with w per 
unit length, supported at the ends. Each of the supporting 
forces is half the total load. The moment about p of ^ivl, at the 
distance p q, is against the hands of a watch, and I call this 
direction negative : the moment of the load w {^l - x) at the 
average distance J p a is therefore positive, and hence the bendiag 
moment at p is 

- \lwl iil -X)- Iw ill - xfY or - {|w;^^ - Iwx^'^ .... (9), 
so that, from (1), ei -j^ = - \ ^wP - ^wx'^ \ . Integrating, we 

P* Q 



:-a-x-^ 



Fig. 230. 



;Wl 



have Ei -^ = - ^w^^x + i wx^ + c, a, formula which enables us to 
find the slope everywhere, c is determined by our knowledge 



that -V^ = where x = 0, and hence e 
dx 



0. Integrating again, 



APPLIED MECHANICS. 409 

Ely = -^-^wX^x^ + ■^-^xvx'^ + c, and c = 0, Lecause v = wheie 
X ~^. Hence the shape of the beam is 

y is greatest where x = \l, and is what is usually called the 

5 w^ X^ 
maximum defl-ection d of the beam, or b = -;r^, — if w = i'w, the 

' 384 EI ' 

total load. 

Example 5. — In any beam, whether supported at the ends or 

not, if iv is constant, integrating (4) of Art. 357, we find 

— — = b + wx, and -^ = a + bx + ^wx- .... (5). 
dx ^ - ^ ' 

In any problem we have data to determine a and b. Take the case 
of a uniform beam uniformly loaded, and merely supported at the 
ends. Measure y upwards from the middle, and x from the 
middle. Then m = where x = \l and - ^l, Q = a -\- \bl + ^wl^, 
and — a - hbl + ^wP. Hence ^ = 0, a = - Iwl^, and (5) 
becomes 

M r= - IwX^ + IWX"- (6), 

which is exactly what we used in Example 4, where wa afterwards 
divided m by e i, and iategrated twice to find y. 

With regard to the following important practical problem : — When 
the sizes of an angle iron are given to find the least radius of gyration of 
its section about a line through its centre of gravity, I give students a 
number of angle irons every year ; for each of them they find the least 
radius of gyration. From the tabulation of these results I hope to get an 
empirical formula. I had hoped to be able to publish this now, but T 
have not yet obtained a sufficient number of results. 



410 



CHAPTER XYIII. 

SOME WELL-KXOWN RULES ABOUT BEAMS. 

340. When beams hare the same section everywhere we 
look for the place where the bending moment is greatest, as 
that is the pLace where fracture tends most to take place, and 
we find the cross section to withstand this greatest bending 
moment. We shall now consider such a uniform beam loaded 
and supported in various ways. 

Thus, the load may be hung from one end of the beam, the 
other end being rigidly fixed, say by being built into a wall. 
When we say that the end of a beam is fixed, we mean that it 
is rigidly held in position, whereas when we say that a beam is 
supported at its ends^ we mean that it is merely held up there. 
In Table YIII. six ways are shown in which the same length of 
beam is supposed to be loaded. The total load is supposed to 
be the same in every case, and the length from a to b is 
supposed to be the same. Then, we see that when the beam is 
fixed at both ends, and the load spread over it, it is twelve times 
as strong as when one end is fixed, and the whole load hung from 
the other end. This means that if, with the beam fixed at one 
end, a load of one ton, hung at the other end, breaks the beam, 
then, when fixed at both ends, and the load spread uniformly 
over it, the same sized beam will carry 12 tons. Hence, if 
experiments are made on the strength of the beam when loaded 
in any of these ways we know what its strength ought to be when 
loaded in any of the other ways. Now a great many experi- 
ments have been made upon beams of rectangular section, 
supported at both ends and loaded in the middle, the third case 
given in the Table ; and from these experiments we know how 
to find the load which such a beam will carry. Having found 
this, we know that when loaded and supported in a different way, 
the beam will carry more or less according to the numbers in 
the column headed " Strength." 

341. If M is the bending moment and z the strength 
modulus of the section, and f the stress which the material 
will stand, M = z/ . . . . (1). 

Let us take as an example beams of rectangular section, 



APPLIED MECHANICS. 



411 



breadtli h, depth d ; the strength moduhis is h d-jQ, so that 

M =/6 (-/3/6 (2). 

Now our theory is based on the idea of perfect elasticity ; 
we cannot, therefore, assume that ar is the bending moinent 
which will break a beam if /is the ultimate tensile or compres- 
sive stress, because our theory cannot hold beyond the elastic 
limit, but we find by experiment that the breaking, bending 
moment is proportional to h d^. Thus if rectangular beams of 
the same material, but of different lengths (7 feet), breadths {h 
inches), and depth {d inches) are supported at the ends and 
loaded in the middle, we find that the breaking load w lb. follows 
with fair accuracy the rule — w r= c & d-jl (3) where c is given in 
the following table and stands for 1/18 of our old,/! It is well 
to try to remember that doubling the breadth of a beam doubles 
its strength, but doubling its depth gives four times the strength. 

TABLE VII. 
Beams Supported at the Ends and Loaded in the Middle. 



Nature of Material. 


c to Calculate Strength. 


e to Calculate Deflection. 


Teak . . . . 


820 


•00018 


Oak 






450 to 600 


•00044 to ^00020 


English Oak 








557 


•0003 


Ash. 








675 


•00026 


Beech . 








518 


•00031 


Pitch Pine 








544 


■00035 


Red Pine 








450 


•00023 


Fir. 








370 


•0005 to ^0002 


Larch 








284 


•00041 


Deal 








GOO 


•00023 


Elm 








337 


•00061 


Cast Iron 








2,540 


•000024 


Wrought Iron 






3,470 


•000016 


Hammered Steel 






6,400 


•000013 


INIarhle , 






150 




Good Sandstone 






50 to 80 





The numbers given in this table are merely the average 
values found by various experimenters. You may wish, how- 
ever, to find for yourself whether they are correct or not. You 
are designing a beam of pitch-pine, say ; then take a rod of 
pitch-pine, 1 foot long, 1 inch broad, 1 inch deep ; support it at 
the ends, and load in the middle till it breaks : the Table says 
that the load will be 544 lbs., but you may iind it to be more 



412 APPLIED MECHANICS. 

or less than this. Remember also that it is near the middle 
that your beam is likely to break ; thi.s^ then, ought to be the 
soundest and most evenly grained part of the timber if possible, 
and the specimens which you try ought to be as nearly as 
possible the same kind of timber. 

If your beam is loaded or supported in any of the other five 
ways described in Table YIII., you will multiply the breaking 
load which you have found by the number called strength in 
Table VIII. The reason is obvious. 

342. In certain standard cases we like to state algebraically 
the amount of bending moment at any section of a beam. We 
shall do this in the six standard cases, so well known to all carpen- 
ters, shown in pages 413 to 415. 

I. Beam of length I inches fixed at one end a, loaded only 
with w at the other end b. At a section x inches from b the 
bending moment is evidently M = w x. The shearing force is 
s = — w. The diagram, Case I., shows m. Notice that m is 
greatest at a, and is then w I. 

II. Beam fixed at a (Case II.,), load w lb. per inch of its 
length or total load w spread uniformly. Note that the load on 
p B, if p is a section x inches from b, would be to x, and the 
resultant of this acts at the distance | x from the section, so 
that its moment {■&\xy,ioxov ^w x^. It is greatest at A, being 
there J ivl-. s the shearing force at p is — wx. Note that 

w 

w I =w, so that the bending moment anywhere is J— cc^ and is 



J w ^ at the end A. The shearing force at the end A is 
numerically greater than anywhere else, being — w. 

III. Beam of length I, load w in middle supported at the 
ends. The supporting forces are each i w. At a section x inches 

from either end the 
1 7 I . bending moment is 

! ■ ! - i w X, bemg - ^ 

,r<wvvvvuMi,'Vivi/VVU/t/U44/VH4yt^^ w^ at the middle, 






t^ O^^-x — ^p I 5^ and the shearing 

(L-x) force s is — ^ w from 

A to the middle, when 

it suddenly changes 

Fig. 231. and becomes + h W 

from the middle to b. 

lY. Beam of length I supported at the ends, load iv lb. per 

inch of its length or total load w spread uniformly {see 



APPLIED MECHANICS 413 

Fig. 231). The supporting forces at a and b are each J w 
or J w I. At p if o p = cc, the bending moment is the moment 
of the supporting force about P minus the moment of the load 
on p B, or, since o b = | ^ and p b = J ? — x, 

-^ ■= — {\wl {\l - x) — w {\l - X) \{\l - x)), 

and this simplifies to m = - -| zt; (?2 _ 4 ^2^^ j^ ig numerically 
greatest at the middle where x is 0, being there — \wP or 
- ;^ w ^. The shearing force at p is ^ l^ ^ — ty (J l—x)^ or w x. 

The student ought himself to draw all the diagrams of jr 
(bending movement) aud s (shearing force). 

The diagrams of m for cases Y. and VI, are shown in the 
figures. (Case VI. will be worked out in Art. 360). They are 
simply the diagrams of III. and IV. with the average value of 
M subtracted from every ordinate — that is, the whole diagrams 
are lowered by this amount. The diagrams of s are exactly 
the same whether a beam is merely supported at the ends or is 
fixed, if the loading is symmetrical (see Art. 362) — that is, the 
fixing does not alter the actual supporting forces at the ends. 

We have, in fact, the following rule for finding the bending 
moment diagram for a uniform beam symmetrically loaded, 
fixed at the ends. Find the diagram of bending moment as if 
the beam were merely supported at the ends : raise it by a 
distance equal to its average height. We now have the diagram 
of the bending moment when the ends of the beam are fixed. 

The shearing force diagram is not altered by fixing the ends. 

If for any two kinds of loading we have the diagrams of m 
and s, then for the two kinds of loading applied at the same 
time we simply add algebraically the ordinates of the separate 
diagrams. 



iU 



APPLIED 31ECHANICS. 













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APPLIED MECHANICS. 



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APPLIED MECHANICS. 



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APPLIED MECHANICS. 417 

343. At the Imperial College of Engineering, in Japan, we 
had a testing-machine with which I made a great many 
experiments with my students. It increased the load on a 
beam at a uniform rate^ and registered the load and deflection 
of the beam at every instant — that is, it drew a curve, each 
point of which showed the deflection and the load which pro- 
duced it. Mr. George Cawley, instructor in mechanical 
engineering at the college, lithographed a number of these 
curves, taken by himself ; and although the experiments were 
made on Japanese wood, so that the actual amounts of load and 
deflection are not of general interest, yet the shapes of the 
curves are so interesting as to be worthy of publication. With 
only one exception, two beams were broken and two curves 
taken for each kind of wood. The mean of these two curves 
has been given in Eig. 232 — that is, a curve lying between the 
two. The specimens were all free from knots. They were all 
28 inches long and IJ inch square. The distance ow repre- 
sents one ton, and the distance o D represents a deflection of 2 
inches, so that the scale of the diagram is known. The load 
was in each case added to at a uniform rate, beginning with o, 
and the rate at which it increased was one ton in two minutes, 
and we see from the figure that practically only in three cases 
did the breaking of the beam take more than two minutes. 
The end of each curve shows where the specimen broke ; it is 
easy to see where the curve ceases to be a straight line — that 
is, where the law, " Deflection is proportional to load," ceases 
to be true ; and this point is therefore the elastic limit. In 
some cases the load corresponding to the elastic limit is less 
than half the breaking load, and in some cases greater 
than this, but usually it may be seen that it is about 
one-half. 

344. What about beams that are not rectangular in section's 
Suppose we have a beam of the same section everywhere, whose 
strength and stiflhess we know, and suppose we want to know 
the strength and stiffness of another beam whicli has the same 
form of section — that is, suppose the new section is such that 
all the old lateral dimensions are increased in a certain ratio — 
then the strength and stiffness increase in this ratio ; if all the 
old vertical dimensions are increased in a certain ratio, then the 
strength increases as the square of this ratio, and the stiffness 
increases as the cube of this ratio. The effect of change of 
length is just the same as it was with rectangular beams, and 



418 



APPLIED MECHANICS. 





























































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iVofg.— students will do well to calculate for each of these materials (1) 
Young's modulus, (2) c and e of Table VII., and insert in an extra page in 
their copy of this book. 



APPLIED MECHANICS. 419 

we know the eff'ect produced by different methods of siippoi-ting 
and loading the beam from Table YIII. 

From Arts. 341 and 342 it is evident that the load which 
a beam will carry without breaking is proportional to the 
strength modulus of its section divided by the length of the 
beam. The deflection of the beam is proportional to the load 
multiplied by the cube of the length, divided by the moment 
of inertia of the cross section. 

345. Beams of uniform strength are those in which the 
nature of the loading is exactly known, and every section is made 
just of such a shape and size as to be equally ready to break 
with all the other sections. There is no difhculty in making z, 
the strength modulus, exactly proportional to m. Thus taking 
up the four well-known cases already described ; let us 
design beams of rectangular section everywhere of breadth h 
or depth d, or of circular section of diameter d, which shall 
be of uniform strength. Note that for a rectangular section 
z a hd"^, and for a circular section z a d-^. In the case of the 
circular section, the plan and elevation of the beam are of the 
same shape. 

Cane 1. — M = wa;, so that hd'^ (x. x. Keep b constant, the 
elevation showing d is a parabola. Keep d constant, the plan 
showing 6 is a triangle, d^ cr x, so that plan and elevation are 
what is sometimes called the cubic parabola ; anyhow, it is easy to 
draw. 

Case 2. — m a x^, so that bd'^ a x-. Keep h constant, the 
elevation showing ^ is a triangle. Keej) d constant, the plan 
showing i is a parabola, d-^ a x'^ ; the plan and elevation are 
easily drawn. 

Case 3. — From the middle to each end, this beam is the same as 
the beam of Case 1. 

Case 4.— M a (/2 _ 4 ^2)^ go that bd^- a [p - 4 x^). Keep b 
constant, the elevation showing d is an elliiDse. Keep d con- 
stant, the plan showing b is two parabolas, d^ oc (^ - 4 a;-), so 
that the plan and elevation are easily diawn. 

We cannot treat Cases 5 and 6 in the same way, because 
we only know m in these cases on the assumption that the beams 
are of the same section everywhere {see Art. 362). 

EXERCISES. 
Rolled girder section, or two equal flanges and web, Hke a, 
Fig. 223. In Tabic VI. we see that the moment of inertia is 

I = j ^# _ {b - bx) d^^ j H- 12, 

and the strength modulus z is i di'sdded by ^d, if. d is the depth 
over all, di the depth betueen the flanges, b breadth of either flange, 
bi the thickness of web. 



420 APPLIED MECHANICS 

1. Ijet the stiTdent calculate i and z in all the cases of the following 
Table, page 421. 

2. Taking / = 20 tons per square inch for iron, and 30 tons for steel, 
find /z in each case. This is the greatest bending moment in inch-tona 
which each section will stand. 

3. Show that a beam I feet long, supported at the ends and loaded in 
the middle, will break when the load in tons is /z -^ 3 I. Calculate this 
for ^ = 10 feet in each case of the table. 

4. Beams 10 feet long. For one ton at the middle the deflection in 

2 240 X 1203 
inches is d = J^ . ^ „s > °^ 2-&%% 4- i. Find this in each case. 

I take E the same for iron and steel — namely, 30 x 10^ lbs. per square 
inch. 

0. An iron beam of the rolled girder section of the table, 18 inches 
deep, 25 feet long, fixed at the ends. What is the breaking load if 
spread uniformly? 

Ans., The load for a 10-foot beam in the table is 89 tons ; for a 25-foot 

89 
beam, supported at the ends, and loaded in the middle, it is — x 10, or 

35-6 tons. Table VIII. shows that fixing at the ends and loading all over 
allows us three times as much breaking load, or 106-8 tons. 

6. What is the mid-deflection of the beam of Exercise 5, fixed at the 
ends, when the load spread all over is 20 tons ? 

Ans., For a 10-foot beam it wotdd be -00223 x -125, according to the 
" deflection " column of Table VIII.; and for a 25-foot beam we multiply 
by 253 -7- lO^, and the answer is -0044 inch. 

7. Compare the strength to resist bending of a wrought-iron I section 
when it is placed like this : I, and like this : m. The flanges of the 
beam are each 6 inches wide and 1 inch thick, and the web is f inch 
thick, and measures 8 inches between the flanges. Ans., ^-bl \ 1. 

8. AVhat is the greatest stress in a bar which is subject to a bending 
moment of 4,000 inch-pounds (1) if the section is a circle of f inch 
radius; (2) if of I form, 2 inches deep and 1 inch wide, the web and 
flanges each being f inch thick. Ans., 5*4 tons ; 3-1 tons. 

9. The dimensions of the section of a cast-iron girder are the follow- 
ing:— top flange, 4 by 1| inches; bottom flange, 12 by If inches; web, 
16 by IJ inches. Determine the position of the neutral axis, and 
calculate the moment of inertia of the section. Find, also, the moment 
of resistance, the greatest permissible tensile and compressive '"tresses 
being 2^ and 7^ tons per square inch respectively. If the girder be 20 
feet long, and is supported at its two ends, find also the greatest safe load 
which it will carry when uniformly distributed along its length. 

Ans., 7|- inch from bottom ; 2,280-5 inch-units ; 800 ton-inches ; 26| tons. 

10. Find the moment of resistance to bending of a beam of wrought 
Iron which has a section like that of the third figure in Table VI., taking 
^1 = 4 inches ; depth, 5 inches ; the thickness of metal everywhere, J 
inch ; and / =20 tons per square inch. What is the greatest load, 
placed at the centre, which a 10-foot beam of this section will stand when 
supported at both ends? Ans,, 60-45 inch-tons ; 2 tons. 

11. In a wrought-iron girder, supported at both ends, the section is 
that shown in the third figure of Table VI. b = 8 inches, ^ = 7 inches, 
d = 10 inches, d = 12 inches. The length is 24 feet. Find the greatest 



APPLIED MECHANICS. 



42J 



•9IPPIH nt nox 

I joj lUBae: ^oo^-ot 

jo saqDui ui uoipayao: 


oooooo999'r^i 


Bending 
Moment of 
Section in 
Inch-Tons. 


w 


lO '^ CM CM (M r-H r-l 


g 


g^S^^^OC.^C.C-1 

cc-cf ^•^^'-^-^^ =^ <^ ^ ^ o 


Breaking Load 
w Tons at 

Middle of 10- 
Foot Beam. 


^• 


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422 APPLIED MECHANICS. 

imifomily distiibuted load which the girder will safely bear, taking 
f T^ 9,000 lbs. per square inch. Calculate the deflection at the middle of 
the beam. ^«s., 12-29 tons; 0447 inches. 

12. A steel plate girder, 70 feet span, 7 feet deep, has a uniform load 
of 1 ton per foot, a detached load of 10 tons at the middle, and loads of 5 
tons each at 15 feet on each side of the centre. Find the diagram of 
bending moment, and state its amount at the detached loads. Calculate 
the best areas of cross-section of the booms at the middle. Take 6 tons 
per square inch in compression and 7 tons in tension as safe stresses. 

Ans., 21-13 square inches; 18-1 square inches. 

13. A hollow tube of wrought iron 20 feet long, 3 inches outside and 
2| inches inside diameter. Find its weight. What is its deflection with 
it& own weight ? What further weight on its middle will it carry safely 
if / = 41 tons p^r square inch ? Ans., 145-2 lbs. ; '44 inch ; 158 lbs. 

14. The supporting forces of a 35-foot beam are 20 and 15 tons. 
There is a load uniformly spread of 10 tons, and one detached load. 
Find the detached load and its position, and find the bending moment at 
the detached load, and also at the middle. 

Ans., 25 tons at 21 feet from one end; 252 foot-tons ; 218*7 foot-tons. 

15. A beam 70 feet long, with weights of 5, 2, 4, 6, and 5 tons at 
distances 10, 20, 30, 35, and 60 feet from one end. Find the bending 
moment at each of these places, and find the supporting forces. 

Ans., 117-1, 184-3, 231-4, 234, and 1029 foot-tons; 11-7 tons; 10-3 
tons. 

16. If we take it that / really means some kind of ultimate stress 
which ought to be considered in the formula for a rectangular section : 

Greatest bending moment =/ . — -, 



we know that in a beam of length I inches, the breaking load at the 
middle being w, the beam being supported at the ends, the breaking 

bending moment must be wi'/4, and hence we have w//4=/-— , or 

J. 72 

yff=lf ,-rL, In the beams of Table VII., h = 1 inch, d = 1 inch, l=V2 

inches. Calculate / for each of the materials mentioned in Table VII. 

17. A beam 12 inches long, 1 inch broad, 1 inch deep, has a deflection 
D for a load of 1 lb. The values of d are given in Table VII. for many 
materials. They are there called e. Calculate the Young's modulus in 

each case. Here d = - becomes 

_ tvl^ _ 1 X 123 X 12 _ 432 
^ ~ 48 I c 48 X 1 X e e ' 

Thus, for English oak e = -0003, or 3 x 10"* and hence e = 1,440,000. 
Divide, therefore, 432 by every number in the "deflection" column of 
Table VII. to find Yoimg's modulus. 

18. A bar of wrought iron 3 inches broad and \\ inch thick is 
supported in a horizontal position at two points 2^ feet apart. What 
deflection at the middle wiU be caused by placing there a load of 15 
c^'ts. ? ' ^«5., 0-0G4 inch. 

19. A straight bar of wrought iron 1 inch x I inch in section ia 



APPLIED JIECIIANICS. 423 

loaded as a tie bar with 5 tons. It is foiuid that the portion between two 
points on it, 4 feet apart, elongates -019 inch. What is the value of e? 
If the bar be subject to a bending moment of 1,800 inch-pounds, what 
would be the radius of curvature? Find also the greatest stress and 
deflection if the bar be supported at points 4 feet apart and loaded with 
120 lbs. at the middle. Ans., 28-3 x 10*^ ; 109 ft. ; 8,640 lbs ; 0-0123 inch. 

20. A square bar f inch x | inch is subjected to a bending moment of 
350 inch-pounds. What is the greatest stress in the bar, and the radius 
of the circle into which it is bent, e being taken at 2,000,000 lbs. per 
square inch? ^«.?., 4,978 lbs. ; 12-5 ft. 

21. A bar of deal 6 feet long, 2 inches broad, and 3 inches deep, 
supported at the ends, is broken by a weight of 1,200 lbs. suspended at 
the centre. What uniformly distributed load would a beam of th; same 
length and material bear if the depth were 4 inches and breadth 3 inches ? 

Ans., 6,400 lbs. 

22. Sketch the diagrams of shearing force and bending moment for a 
uniform beam, supported at both ends, 30 feet long, weighing 10 lbs. per 
foot run, and carrying a load of 1 ton at a point 12 feet from one end. 
Give the numerical values of the shearing force and bending moment for 
the middle transverse section of the beam. 

Ans., 896 lbs. ; 174,780 inch lbs. 

23. A beam 42 feet span supports five wheels of a locomotive. The 
fore wheel is 1 foot from the left end, and the distances between the 
wheels, in order, are 5, 8, 10, and 7 feet, and the loads transmitted to the 
beam, in order, are 5, 5, 11, 12, and 9 tons. Find the maximum bending 
moment. Ans., 3,612 inch tons. 

346. In designing railway girders the theory of bending is 
found to lead to useful rules ; but in machine design some judg- 
ment is necessary in applying rules, and want of judgment is 
conspicuous sometimes in the writers of books on this subject. 
What we would urge upon students is the necessity for a great 
respect being shown to what are sometimes called "rule-of- 
thumb-proportions " : great respect, tempered by criticism. 
When all the ])eople of an old trade have made a pedestal or 
hanger of much the same proportions, we must remember that 
these proportions have been reached at considerable cost in 
trials and failures. 

347. The Teeth of Wheels.— When toothed wheels drive 
each other, their teeth tend to break like little beams fixed at 
one end. It is usual in considering their strength to regard the 
pressure between two teeth as acting at a corner, because this 
may accidentally occur, and it is the most trying condition. 
There are usually two pairs of teeth in contact at once, so we 
consider that only half the total horse-power has ever to be 
transmitted by one pair of teeth. This transmitted horse-power, 
multiplied by 33,000, divided by the circumferential velocity of 



424 



APPLIED MECHAyiCb 




Fig. 233. 



the wheel in feet per minute, is, of course, the force in pounds 
which each tooth has to withstand (see Art. 41). Imagine the 
tooth to tend to break at a section ai. c d b, Fig. 233, making 45" 
with the depth a H, just as we know- 
it would break if the corner were 
fe-truck smartly with a hammer. This 
consideration leads to the rule that 
the square of the pitch is propor- 
tional to the force, divided by the 
greatest safe stress per square inch 
to which the material may be sub- 
jected. If teeth are so carefully 
trimmed up that we imagine the load 
always to be distributed over the 
whole breadth, it is easy to see that 
the pitch ought to be proportional 
to the force per inch breadth of wheel. Taking the first 
case, it is easy to see that the practical rule becomes 

p = c '/b. V 

where h is the horse-power transmitted at a velocity of v feet 
per minute. If jy is the pitch in inches, c may be taken as 7 for 
well-made cast-iron wheels working without shock ; c is 9 for 
ordinary mill wheels ; c is 11 in wheels subjected to shocks ; c 
is also usually taken as 1 1 in mortise wheels. At high speeds 
c is usually taken gi'eater as shocks are probably gi-eater. 

EoyiYcwe. — A sjjur wheel making 100 revolutions per minute 
has 60 cogs of 3 inch pitch (the circumference is therefore ISO 
inches, or 15 feet, so that v ^^ 1,500 feet per minute); what 
power will it transmit safely? Here vp^jc- is 1,500 x S-r-c^, so 
that the horse-power is — 

275 in the most favourable case. 

167 in mills, 

112 when there is much shock. 



"When teeth are well trimmed we may take it that the above 
numbers fore ought to be diminished by 25 percent, if the breadth 
of the wheel is four times the pitch. In very broad bevel wheels 
the above velocity is to be taken as the average velocity. In 
ordinary or narrow bevel wheels it is the velocity of the inner 
parts of the teeth. 



APPLIED MECHANICS. 



425 



348. Similar Structures Similarly Loaded. — If a girder is 
loaded mainly by its own weight, then any other girder made to 
the same drawings but on a different scale would be a similar 
structure similarly loaded ; and this is the name given to all 
structures made from the same drawings but to different scales, 
if their loads are in the same proportions to the weights of the 
structures themselves. It will be found that in all such cases the 
stress at similar places is proportional to the size of the structure 
— that is, the weakness of the structure is in direct proportion 
to its size. 

This is easily seen if we imagine the structure to be such a 
simple one as a rod, A, Fig. 234, carrying a weighty ball, w. If 
there is another such arrangement, of twice tlie size in every 
direction, the area of cross section of the rod would 
be four times as great ; but the load to be carried 
would be eight times as great, and therefore the 
stress per square inch at a section would be 1 wice 
as great — that is, the larger rod and ball would be 
twice as weak. As the stress would be twice as 
great and the length of the rod twice as great, the 
extension would be four times as great. The exten- 
sion of the rod per foot in length would only be 
twice as great. In the same way a beam of cast 
iron, 1 inch square and 1 foot long, is 1,700 times 
too light to break with its own weight, wheieas a 
beam of cast iron whose length, breadth, and depth 
are in the same proportion, if 1^700 feet long and 
1,700 inches square in section, would break with 
its own weight. The deflection of similar beams similarly 
loaded is proportional to the square of their dimensions ; but 
the deflection per foot of length is only proportional to their 
dimensions. 

Imagine similar beams of the same material whose similar 
dimensions ai-e as 1 to s and the loads are as 1 to s^ ; evidently 
bending moments will be as 1 to s*; moments of inertia of 
cross sections will be as 1 to s* ; curvatures at similar places 
will be equal ; deflections will be as 1 to s^ ; stresses will be as 
1 to s. 

If we dare take the loading of similar ships as 1 to s" where 
n is something between 2 and 3, bending moments will be as 
1 to s^ + ^ ; curvatures will be as 1 to s^~^ ; deflections will be 
as 1 to s°"^ : stresses will be as 1 to s°~^* 




Fig. 234. 



426 



APPLIED MECHANICS. 



In machines, accelerating forces are proportional to S X 
masses x N^ if n is the number of revolutions per second or per 
minute. Hence in similar machines the forces are proportional 
to S-* n2, bending moments are proportional to s^^ N^, curvatures 
to s n2, deflections to n^ s^, stresses to s^ n^. If, therefore, the 
stresses due to mere accelerations in machines are to be the 
same, s n must remain constant — that is, if the size of a machine 
is doubled its speed must be halved. 

348a. Strength of Ships. — Let a c represent the length of a 
ship. Imagine the ship divided hy sections at equidistant 
points in A c. Let the ordinates of the curve "W represent 
the weights of these portions, and the ordinates of the curve b 
represent the buoyancy or weights of water displaced by them. 
The two areas must he equal, and their centres of gravity lie on the 
same ordinate. The ship is " water-borne " at d and e. We take 
the vertical distances between a b d^ and a w d^, and between c b e^ 
and c WE^, as representing the downward load per foot or per inch 




on abeam a d, and on c e and the vertical ois canoes between d* b e^ 
and D^ w E^ as upward load per foot or per inch on the part of the 
beam db. If we set ofE these distances, therefore, as ordinates 
from a line a c (Fig. 234a), we have the carve of positive and 
negative loading of a beam. "We can use them to obtain the shear- 
ing force and bending moment at every section of the beam or ship. 
Even when lying level, it is evident that the loads on a ship, 
regarded as a beam, are in other directions besides the vertical. 
Shipbuilders must also take account of the loading due to the 
inertia of the parts of the ship in her various kinds of motion. 
When these loading forces are known it is not difl&cult to calculate 
the strength. Besides the strength as a whole, it is important to see 
how each part communicates the load on it to the general system. 
The complete problem is therefore a complicated one, which, how- 
ever, can all be worked out according to the principles given in this 
book. The general student is supposed to know the above simple 
method of obtaining loading and, therefore, bending moment and 
shearing force at each section of the ship as if it were a beam. 



427 



CHAPTER XIX. 



Fio;. 235. 



DIAGRAMS OF BENDING MOMENT AND SHEARING FORCE. 

349. When a beam is loaded, in any way whatever, it is easy to 
obtain by a graphical method the diagram of bending moment ; 
in fact, in finding tlio supporting forces one finds the diagram 
of bending moment. Let A b (Fig. 235) represent the length of 
a beam which has three ver- 
tical loads— 1, 2, 3. To find 
the vertical supporting forces 
at A and b, draw the unclosed 
force polygon, k l (Fig. 236) — 
before the student arrives at 
this part of the book he will 
probably have drawn other 
force polygons where all the 
sides were really in the same 
straight line — 1, 2, a*nd 3 (Fig. 
236), representing in direction 
and magnitude the three loads 
of Fig. 235. Choose any point 
O. Join o K, o 1 2, o 2, 3, and 
o L (o 2, S, means the line join- 
ing o with the point where the 
sides 2 and 3 meet). Now draw 

the link polygon (Fig. 235), ■• 

beginning at any point a, in the vertical from A, and ending 
in the point b. Now ah is the side closing the link polygon. 
Draw ON (Fig. 236) parallel to ab (Fig. 235). Then l n is 
the amount of the supporting force at b, and n K is the 
amount of the supporting force at a. Also draw any vertical 
line, s T (Fig. 235). Then the length s t, intercejoted by the 
sides of the link polygon, represents the bending moment of the 
beam, at any point p on some scale which it is easy to find. 

To prove this : Draw o h horizontally. The moment at any 
point p due to the supporting force, n k at a, is n k x a p ; and 
this is equal to o h x ft; for, by similar triangles, 

aT:Tr::oN:NK, 
and therefore a p : i' r : : o h : n k. 



Fig. 236. 



428 APPLIED MECHANICS. 

This second proportion gives x k . a p equal to o h . t f. In the 
same way tlie moment at p due to the force 1 is o h . p s ; £ind 
hence the true moment at p, being the difference of these, is 
OH . ST. Let the student prove as an exercise that if the beam is 
di-awn to a scale of 1 inch represented by ^' inches, and if the loads 
are ditiwn to a scale of 1 pound represented by i/ inches, then s t 
is the bending moment at p, on a scale such that 1 poimd-inch is 

represented by — ^ inches, o h being measured in inches. 

If the load is not concentrated at a number of points, it is 
usual to imagine it divided into a ntunber of loads, each of which 
acts at one point. The diagram of bending moment is drawn in 
the way which I have just described, and then for the polygon 
with its straight sides we substitute a curve which touches all the 
sides of the polygon. 

After you have found a diagram of bending moment, if you 
wish to see the effect of additional loads, draw a diagram for these 
loads a6 ii they acted alone, but take care that the horizontal 
distance o h is the same as before. Add together the ordinates of 
your two diagrams to get your new diagTam of bending moment 
for all the loads. 
350. It is very unfortunate that this subject is usually 
taken up from an academic standpoint. Students discuss much 
theory sufficiently well for examination purposes, but they do not 
understand the most elementary things. They seldom state on 
a drawing what is the actual bending moment in pound or ton- 
inches at any section. Kor, indeed, do they ever seem to 
use these drawings for practical pui'poses. Instead of one 
diagram covering a large sheet of paper, we have the usual 
diagram looking like a mere book illustration in size. 

A student wdll find that unless he works a number of 
exercises graphically he cannot comprehend this subject. Let 
him take such an exercise as this. A beam a b is loaded with 
5 tons at c, 7 tons at D, 2 tons at E where A c is 8 feet, a d is 
13 feet, A E is 18 feet, a b is 24 feet. It is also loaded with 1 
ton per foot from a U c, 0'5 tons per foot from c to d, 0*7 tons 
per foot from D to e, and 0-8 tons per foot from e to B. It is 
worth while making a diagiam for the distributed load, its 
ordinate showing tlie amount of load per foot or per inch. 

Take also a case with no 
detached loads but only a 
diagram showing iv lb. per 
i i . I inch as shown in Tig. 237. 

'^ ^- We divide the area up into a 

I convenient number of parts, 
Fig. 337 ' c F H A, F I K H, ifec. Find the 



E 



APPLIED MECHANICS. 429 

centre of gravity of u L K and let tlie arrow there represent 
the area i J l k — that is, the load on the portion k l. The 
student ought to state to what scale his area represents 
load. Thus we represent the whole by a number of detached 
loads; we draw the diagram of bending moment which is a 
polygon, but it is to be noticed that lines f h, i k, j l, &c., pro- 
duced, meet the polygon at points which are on the true bending 
moment diagram for the distributed load, and so the curve is 
easily drawn touching the polygon at these points. 

351. It is proved in Art. 339 that if d is the greatest distance 
moved hy any point in a heam, and this is called the heam's deflec- 
tion, and if the cross-section is the same everywhere, w the load, 
L tlie lengtli, i the moment of inertia of the section, and e the 
modulus of elasticity, 

D = ~ for a beam fixed at one end and loaded at the other. 

3 EI 

D = r — for a heam fixed at one end and loaded uniformly. 
8 E I •' 

1 w L^ for a heam supported at the ends and loaded in 

^ ~ 48 eT* ^he middle. 

_ 5 w L^ for a heam supported at the ends and loaded 

^ ~ 384 eT uniformly. 

The third of these formulae is the one most needed. It is by 

means of this formula that the modulus of elasticity is generally 

determined. Thus in careful experiments with an iron beam 

1 inch broad, 1 inch deep, carried on supports 24 inches asunder, 

suppose we find that a load of 2,000 lbs. produces a deflection of 

1x1x1x1 

one-quarter of an inch. Now, i for the beam is 

1 

or — . The third formula given above becomes 

_ 1 2,000 X 24 X 24 X 24 
16 3 E X tV 

and from this we find that e is 27,648,000 lbs. per square inch. 

Again, taking the third of the cases shown, I find that 560 lbs, 
produced a deflection of 0-22 inch in a beam of wood 24 inches 
long. If inch square, supported at the ends. Here 

I = 1-75 X 1-75 X 1-75 X 1-75 ~ 12, or -781, 

, „_ 1 560 X 24 X 24 X 24 

and -22 = — — , 

16 3e X -781 ' 

from which we find that e is 938,656 lbs. per square inch. 

Agaia, from Table VII., we see that a beam of teak 12 inches 

long, 1 inch broad, 1 inch deep, gets a deflection of -00018 inch for 

a load of 1 lb. Here the moment of inertia of the cross-section 

is ~ and -00018 = i ^ "" ^e^'x^J-'' ^^ ' ^""^ "^^^^ ^^ ^^ 
that E for teak is 2,400,000 lbs. per square iach. 



430 



APPLIED MECHANICS. 



352. Take a small beam,, a b. Fig. 238, supported at the ends, 
•-^nd load it in the middle. Measure carefullv the deflection or 
lowering of the middle point This is called the deflection of 
siuch a beam. Now this distance will usually be small, and so 
we had better magnify it by letting the string cw pass over 
the little axle e. which carries a long pointer. This pointer wiU 
^how on the scale p k a magnification of the deflection. We 




Fig. 238. 

shall find that the more load we place at c the greater is the 
deneetion : and in fact the deflection is proportional to the load, 
until our loads become great enough to produce permanent set, 
when (Art. 244) the deflections increase more rapidly than the 
load. If now Ave use a beam of the same material bnt of double 
the bi*eadth, then for the same load we shall get one-half the old 
deflection. If we use a beam of double the depth, then for the 
same load we sliall get only one-eighth of the old deflection. 
Also, if we double the length of oui* beam, using the same load, 
we shall get eight times the old deflection. A very instructive 
series of experiments may be made very easily in this subject, 
and we shall not thoroughly understand the matter unless we 
make a few such experiment.s. It is found that a beam of pitch 



APPLIED MECHANICS. 431 

pine, 1 foot long, 1 inch broad, and 1 inch deep, supported at 
its two ends and loaded in the middle, is deflected -00035 inch 
by a load of 1 lb. This explains the numbers given in Table YII. 
It is found that if the same beam is fixed at one end and loaded 
at the other (first case of Table YIIL), the deflection is 16 times 
as great, whereas if the beam is fixed at both ends and the load 
is spread uniformly (last case of Table VIII.), the deflection is 
only -125, or one-eighth as great. This explains the "deflec- 
tion " column of Table YIII. 

The rule, then, to fmd the deflection in inches of any beam 
loaded in any of the ways shown in Table YIII. is this : — Multiply 
together the cube of the length in feet, the totalload in pounds, the 
number called deflection in Table YIII., and the number called 
deflection in Table VII., and divide the product by the breadth of 
the beam in inches, and by the cube of the depth in inches. 

Example. — A beam 20 feet long, 10 inches broad, 15 inches 
deep, of pitch pine, fixed at one end and having spread all over 
it a total load of 4,000 lbs. — what is its deflection ? Here the 
number in Table YIIL is 6, and in Table YII. it is '00035 ; hence 
we have 20 x 20 x 20 x 4,000 x 6 x -00035 divided by 10, and 
again divided by 15 times 15 times 15, which gives as answer 1-99 
inch. The end of the beam would be deflected this distance. 

A beam is said to be stiff if its deflection is small, and we 
say that the stiffness of a beam supported and loaded in 
the various ways shown in Table YIII. is for the various cases 

— , ^) 1 5 1 *6, 4, 8. In fact, a beam of a certain length carrying 

a certain load is 128 times stiffer when it is fixed at the ends 
and loaded uniformly than when it is fixed at one end and 
loaded at the other end. 

It is well to remember that when we double the breadth of 
a beam we double its strength and also its stiflhess, but if we 
double its depth we get four times the strength and eight times 
the stifiness. Beams required to be very stiff ought to be very 
deep. Care must be taken, however, that they are laterally 
supported, else they will buckle. If you double the length of 
a beam you get half the strength, but you only get one-eighth 
of the stiffness. 

353. The student must see that the heights of the points 
on the force polygon ; for example, the heights of the points 
between 1 and 2, 2 and 3, &c. (Fig. 236), above N, give him the 
)rdinates of his shearing force diagram so that he can obtain 



432 



APPLIED MECHANICS. 



this diagi'am by horizontal projection. When the loadvS are 
distributed, at points like h, k and l of Fig. 237, the ordinates 
are the same as if the loads were detached, and they may be 
joined by a curve. 

We shall now consider a number of cases where arithmetic 
and algebra help out our graphical methods of working. 

354. If tlie loads on a structure are Wj, Wg, etc., the stresses and 
strains everywhere are the sum of those that would he produced if each 
load acted alone. "We find this true in all cases that we try, unless, 
indeed, in certain cases where instahUity is produced. It is usually 

supposed to he a mere 
statement of the mathe- 
matical law of superj)osi- 
tion of small effects. We 
know, at all events, that 
the hending moment at 
any pa^rt of a heam due 
to loads w^, W2, etc. , act- 
ing together, is the sum 
of the hending moments 
due to each acting singly. 
Exercise.— InFig. 239 
loads R = 10 tons and s = 
15 tons act at the ends c 
and D ; supporting forces 
p and Q act at a and b. 
Draw the diagrams of 




Fig. 239, 



c A = 5 feet, AB = 15 feet, b d = 3 feet, 
bending moment and shearing force. 

Answer. — ci:fdc is the diagram of hending moment, where 
AE represents 50 foot-tons and bf represents 45 foot-tons. The 
supporting forces, a\ hich are easily found graphically and analytic- 
ally, are p = 1033 tons, o, = 14-67 tons. Hence the diagram of 
shearing force is dghijklcd, where dg= -15, bi= - 0-33, 
AK. = + 10, all in tons. 

Exercise. — In Fig. 240 a weightless heam rests on supports 





^ 


p 


Fig. 240. 


Q 


^ D 



at A and b ; torques m-^ and ?«2 are applied at c and d. Find 
the hending moment and shearing force everywhere. 

Evidently p and o, are equal and opposite ; p x a b = Wi - Wg, 

^ 2„_ 1 ^■''^pisan upward force, 



or p = 



a = - 



If 



AB AB 

Q, must he a force holding the heam down at b. Thus, let 
Wj = 30 ton -feet, m^ = 17 ton-feet; let ab he 15 feet. Then 



APPLIED MECHANICS. 



433 



p = 13/15 ton, Q = - 13/15 ton. s is zero from d to b, - 13/15 
ton from B to A ; and again zero from a to c. The bending 

moment from d to b is 
^ " constant and equal to m^ ; 

from B to A it increases 



stant from A to c. 

CGHjKDcis the dia- 
gram of bending moment, 
where c g = a h represents 
m-^, and j b = k d = m^^. 
DBEFACD is the dia- 
gram of shearing force 
where b e = a r = - 
13/15. 




c 






D 


' A 


\, 


T y 


B 
■Q 




N \ 





Fig. 241. 
Exercise. — Let the beam of Fig. 240 hare moments m^ and m^ 

applied at its ends, as shown in Fig. 241 ; also let it have a load of 

Wi lbs. (say 5 tons) at t (say that at is 6 feet). This of itself 

would produce bending moment, as shown at alb of Fig. 242, 

where t l is - 18 ton-feet, ^ ^ 

and would produce support- 
ing forces p = 3 tons, ci = 2 

tons. The shearing force 

diagram is bxvtunab, 

where b x = t v = 2 tons, 

and AN=TU= - 3 tons. 
Now imagine the beam 

loaded with 1 ton per foot 

on c A and '5 tons per foot 

on B D. Jf c d' = a;, the 

bending moment at d' is 1 > 

or 12-5 ton-feet. The bending moment curve ce is parabolic. 

Similarly the bending moment curve d f is parabolic, the moment 

being i x 0-5 x 3^ or 225 
ton- feet at b, shown at b f. 
The bending moment due 
to the loads on the ends 
would be c E F D over the 
whole beam, and the sup- 
porting forces would be 
p == 5-6833, Q = 0-8167. 
The shearing force dia- 
where bi — - 1^ tons, bj = ak 



Fig. 242. 
I a; or i x^^ and at a it is | (c a)^ 




Fig. 243. 



gram is dijkaecd (Fig. 245) 
= - 0-6833 ton, ae = 5 tons. 

Now imagine the beam to have a uniformly distributed load 
of \ ton per inch be- 
tween A and B. The sup- 
porting forces required 
by this are p = 2-5 and 
a = 2-5 tons. The bend- 
ing moment diagram is a E 
parabolic curve a e b, where Fig. 244. 









&' 


A 


"5.--^^""^ 


B , 


V 


[>< 


^ ^^' 


■a 



434 



APPLIED MECHANICS. 



DE = - 9-375 ton-feet. The shearing force diagram is B D h a B 
(Fig. 244), where bg = 2-5 tons, ah = - 2 "5 tons. 

Let the student work ^ 

out all these diagrams 
and add all the ordinates 
together. 

Xow let him obtain 
the same result by one 
graphical construction. 
The continuous loading 

he will break up into a number of detached loads, and the following 
example will show him how to work the problem. Let the beam 
KL N p be loaded and supported as shown in Fig. 246. Use Bow's 
lettering. 

355. In Fig. 246a the loads are given, half -barbs being used for 



245. 



^ 




Rg. 246. 

the arrow-heads ; the missing comer of the force 
polygon ABCDEFGHi is I. Choose a pole 
and join 




Fig. 241* 



and join with 
ABC, etc. We can- 
not yet draw 1. Now 
start at k, or any point in the line of 
load AB (Fig. 246), and draw a Hue 
through the space a parallel to a (Fig. 
246a) to meet (produced) the force a i 
in Q. Also draw k x through the space 
B parallel to b to meet the load-line 
B c in X. Draw x w through c pai-aUel 
to c, and so on to r. Now join k and 
a. This is parallel to the missing Hne 
I (Fig. 246a), which may now be drawn 
and I f oimd. h i and i a (Fig. 246a) 
are the amounts of the supporting 
forces. The diagram qaxwtutsrq 
is the diagram of bending moment, 



APPLIED MECHANICS. 



435 



vertical distances such as vy representing the "bending moment 
at each corresponding section. The scale of measurement may 
he computed by Art. 349. "We have positive bending moment 



Fig. 247. 

(tending to make the beam convex upwards) between k and z 
and between m' and p, and we have negative bending moment 
between z' and m'. z' and m' are places of no bending moment, 
where the beam has no curvature ; they are called points of 
inflexion. The shear force diagram 1, 2, 3, 4, 5 presents no 
difficulty. 

It is nsual, when we finish the work, to draw the diagram of 
bending moment as in Fig. 247, the ordinates being measiu-ed, not 
from a broken line as in Fig. 246, but from a horizontal hne. 

Travelling Loads. 
356. I. Suppose the load w (Fig. 248), to travel over the beam 
from A to B. "When w is at any point c, between a and d, the 






^^^^^f^^ 




shearing force at d (say Sd ) is positive, and equal to the reaction q. 
This positive shear at d increases with a as the load approaches d, 

and in the limit, when w is very near to d, it has the value — x. 

At the instant the load passes i>, the shearing force at d 

diminishes by the amount w, and becomes - x - w, or - - (l - x), 

thus becoming negative, and equal to the reaction at p ; and as the 
load moves on towards b, the negative shear at d diminishes 



436 



APPLIED MECHANICS. 



numerically. We thus see that the greatest positive shearing force 
at any place occurs ^-hen the load is just to the left of the place, 
and the greatest negative shear when the load is just to the right. 

Expressed algehraically :— Max. + Sd =— x; max. - sd = -~(l-x). 

These are the equations to straight lines, and the corresponding diagrams 
are set out in Fig. 248 :_ that for maximum positive shear is the triangle 
ABE, and that for maximum negative shear the triangle b a f. 

Next consider the maximum bending moment at d (sav max. Md). 



^t^^ 




As the load advances from a towards d, Md increases, since its value is 
q (l -x), and a increases. When the load passes d, Mi, diminishes, since 
its value is now p x, and p diminishes. TTierefore the maximum value 
of Md occurs when the load is at D, and its value is given by the equation 



max. iiT, = p 



or a (l - x) = w 



This is the equation to a parabola passing through a and b, axis 
vertical, the ordinate of the vertex [see Fig. 248) being w l/4. 

II. Two travelling loads Wj and Wg at a fixed distance / apart ; 
draw the diagrams of maximum positive and negative shearing 
force and maximum bending moment. 

As the front of the first load Wi approaches Dj + Sd increases. 



APPLIED MECHANICS. 437 

since its value = q ; when v,\ passes d, s d undergoes a sudden 
diminution by the amount Wx ; and as w^ moves from d towards b 
the value of + s „ increases, on account both of the advance of Wi 
towards b and the approach towards d of the other load Wo. This 
increase goes on until Wj crosses d, when there is a second sudden 
drox5 i^i the value of + s n, followed by a gradual increase until w\, 
arrives at b. We see that the maximiun positive or negative s*^ 
occurs when the front or back respectively of Wi or Wg is at d. 

To draw the diagram : Draw the diagTams of s for the front 
and back of each of the loads Wj and Wo in its passage across the 
beam ; the boundary line is the diagram required. The diagTam 
for fi'ont of "Wi will be a straight line through a, the equation of 

which is + s = Q = Wj -, until Wg comes on, when the slope 
suddenly alters, the s at front of w^ then being 

X X - ^ , , X I 

+ S = Q = Wi - -h W^2 — = (Wi + W2) - - W2 -. 

Li Li Li Li 

Wl + W, 

The slope now is '-, and the diagram of s in fron'j of Wj 

continues a straight line in this direction. Next, for s at back of W]. 
This is equal to s at front - "VTi for all positions, and the diagram 
is a broken line parallel to the one just di-awn at a distance below 
it, equal to Wi. Next, for s at front of Wg. The s between "Wi and w^-, 
is the same at all points ; that is, when W2 is just coming on, the 
s in front of it is the same as the s at back of Wj. Through k 
draw the horizontal line Jc m, then m is the starting-point for the s 

diagram in fi'ont of w^. The initial slope is — -^ : this extends 

to a point distant horizontally I from b, when w"i passes off the 

beam, and the slope diminishes to — . The diagram for the rear of 

Wg is parallel to the line just drawn and at a distance below it, 
equal to W2. As a test of accui-acy, it is to be noted that this broken 
line should end at b. 

Consider now the maximum bending moment. Let aVj + Wr 
at t be the resultant of w^ at a and Wg at b. Let ta = a and 

tb = /3. Then - = — , that position being shown in Fig. 249 
P ^1 

in which the loads w^i and W2 are one on each side of d. Let the 
distance of line of load w^ from d be x. The maximum Md e\a- 
dently occurs either when Wi or Wo is at d, or for some intermediate 
position. Consider an intermediate position as shown. Md = 

Q (l - X) - W^X = (Wi + W2) ^ "^^ ~ " (l - X) - -W^X = (Wj + W2) 

^ ° (l - x) + a; { (Wj + W2) ^^-^ - Wi j . It will be seen that 

so long as the expression in the last bracket is positive, Md is a 
maximum when x is greatest — that is, when x = ly or, that is, when 



438 



APPLIED MECHANICS. 



L 



Wj [ is positive 80 long as 



When 



IS > 



w, + w, 



^, tlie expression in 



load Wg is at d. And 

L Wi + W2- 

L - X Wj 

tlie last bracket is negative, and the value of Mq is therefore a : 
mum when x is zero — that is, when load Wj is at d. 

To draw the diao-ram, take point t in a b such that — = — • 
° ' ^ BT M-j 

Let AT be called the field of Wg, andfi t the field of Wi; then Md is 

a maximum for any point d in the field a t when the load w^, which 

governs that field, is directly over point d, and for any point of the 

field B T when w^ is at the jpoint, provided in each case the load can 

completely traverse its field without the other load leaving the beam, 

which condition requires I to be not greater than the smaller of 

the two values — l, or — l. 

Wi + W2 Wi + W2 

Curve for field b t. Putting a; = in the expression for Mjj, we 



have maximum iiD = 



Wi +^2 



(x - a) (l -x), which is the equa- 



tion to a parabola passing through b and a point n distant a from a; 
and by symmetry the curve for field at is a parabola passing 
through A and a point k distant (3 from b ; the two parabolas will 




Fig. 250. 
be found to intersect at i directly under t. The distance x of the 

mid point of b n from a is a H — = — - — . Putting this ^-alue 

in the equation to the curve, we have the ordinate h y = -^ ^ 

\h+l - a] [l - "4^1 = ^^ii^(L - a)- the correspond- 

ing value for g x, or depth of other parabola a k, is g x = — 

(l - jS)^. When the distance I is greater than the shorter of the 



APPLIED MECHANICS. 



439 



two fields, there is then a third parabola through A and b coiTe- 
sponding to the greater of the two loads taken alone by method 
shown in Case I. 

III. A load as of a travelling train, to lb. per unit length, comes 
upon a girder a b from the left, covering it from end to end, and 
then leaving it. Show that the greatest positive shearing force at 
a section d occurs when the front of the train reaches p, and that 
the greatest negative shearing force occurs when the rear of the 
train leaves d. Also find the maximum bending moment at d. 

We have seen that any load produces positive s if it is to the left 
of D, and negative s if it is to the right of d. Hence, to produce the 
greatest positive s at d, there ought to be no load to the right of d, 
and for gTeatest negative s at d there ought to be no load to the left of 
D, so the proposition is proved. When the train covers a d, s at d = a, 



2l 



. (1), being shown by d g in Fig. 250. When the 

. (2), being 



train covers d b, s at d is - p, or 



2l 



(L - x)2 



shown at D H in Fig. 250. b e and f a are numerically equal to half 
the load when it covers the span. The curves ae and be are 
parabolic, the equations of which are given in (1) and (2) respect- 
ively. The student ought now to add to the ordinates of Fig. 250. 
the shearing force due to a uniformly distributed constant load, 



N\\^^^^^ 



5#^ 




Fig. 261. 



and write out in words actually what s is as a train rolls on, covers, 
and rolls off the bridge. 

Next consider the maximum m at d. Any load anywhere on a 
girder increases the bending moment anywhere. Hence where 
load due to a travelling train comes upon a girder, covers the girder, 
and leaves it, the bending moment at any place is never greater 



440 APPLIED MECHANICS. 

than what it is when the whole girder is covered. Therefore, the 
maximum bending moment will occur when the girder is fully- 
loaded. A curve to this is known to he a parabola through a an<^ 

B, the depth of vertex being -3-. 

o 
IV. Travelling load of uniform intensity w lb. per unit length, 
of length I less than the span l. The maximum positive s occurs 
when the front of the load is at d, and maximum negative when the 
rear of the load is at d ; so that while the load is only partially on 
the beam the diagi'am of maximum + s will be a pai-abola similar 
to that of Case III., the equation being for a distance I from a 

maximum + s = -„— • When the load comeswhoUyon the beam, as 
in Fig.251, the diagram alters ; maximum + s = q= — (x-^) or 



w 



( X - - V The equation to a straight line slope -, intersecting 

A B at a distance - from a, and therefore tangential to the parabola. 

The diagram of positive s is most readily set out by first drawing the 
line E c, as shown in figure, then drawing the parabola to touch it 
at E, and the line a b at a. A similar curve set out downwards 
fiom B will be the diagram of maximum negative s. 

Maximum bending moment at d. It is easily seen that the 
maximum Md occurs either when the front a or the rear b of the 
load is at d, or for some intermediate position, as in Fig. 261. To 
find value of m d for an intermediate position let the condition of 
the co-ordinates be as in Fig. 251. 

M^ = a (L - X) - — = - [(x + ^ - - j(L - x)J - — 
For a maximum, -j^ = (x being constant), .-. maximum m d occurs 

^ wl . . - L-X X . , BD ait 

when — (l - x) - wa; = 0, or = , ; i.e. when — = ^— , 

L^^^ X I - x' adJd' 

or the maximum m d occurs for such a position of the load that d 
di-^-ides it and the beam into segments, the ratio of which are 
respectively equal. 

Putting this value of x, namely x = I. , into the expres- 

sion for Mp, we have maximum Md 

__WIV ^(L-X) n Wl\-L - X)^ 

equation to a parabola passing through a and b, the depth of the 
vertex being 



WL 

"4" 



(^ \ W L W I. , ^, 



441 



CHAPTER XX. 



MORE DIFFICULT CASES OP BENDING OF BEAMS 

357. We do not usually trouble ourselves as to whether we call the 
bending moment which makes a beam conyex upwards positive or 
negative, representing it by upward-drawn or by downward-drawn 
ordinates. There is the same sort of choice in regard to shearing 
force. But for the sake of ha^dng plus signs in the following 
expressions (1) and (2), w^e had better adhere to the following- 
definitions. A section which is being sheared is supposed to be at 
the positive distance x to the right of the zero point. Loads are 
positive forces ; supporting forces are negative. Positive shearing- 
force s means that the material to the left of a section acts with 
downward force on the material to the right of the section. 
Positive bending moment m causes a beam to be convex upwards ; 
positive 2/ at a place is a downward disDlacement. When this is 
the case we know that 



dec 



i . . . . (1), 



di d^t/ 

dx dx^ 



.... (2). 



In the same way we can show that 

d s 



i^ = s....(3), 



dx 



- w .. .. (4). 



To prove (3) and (4), consider the equilibrium of the portion of 
beam between the sections A b and on. At a b there is the bending 
moment m and shearing force s, and 
at CD there are m -f 8 m and s -f 5 ?, 
and 0' is 5 x. The forces acting on 
this portion of beam are shown in 
Fig. 252. The load being w per 
unit length, the resultant load here 
is w .'^x. Hence, considering- the 
vertical forces, S s — w . dx ovd s/dx 
— w . . . . (4:). Taking moments 

about 0',u-\-s.Sx + ^w {Sx)'^ = m+ 

5 M or S u/5x = s -{- ^w .Sx; and in 
the limit, as 6x gets smaller and 

smaller, -7— = s . . . . (3). 

358. We saw (Art. 349) that if we 
have a diagram of tv we can find easily, 
graphically, the diagram of m. We 
now see that if the value of m at Fig. 252. 

every section be divided by the value 

of E I there, and if we treat this diagram, showing m/e i every- 
where, exactly as we treated the w diagram, we obtain y. 



A 


^ 


.^U 




._-._^-_.^, 




~o 


0' 


B 


D 







442 APPLIED MECHANICS. 

This gTapliical metliod of -forking is quick and accurate. If we 
only possessed an accurate mechanical integrator, such that when 
given a curve showing x and i\ v heing a function of x, we could 
at once draw another curve whose ordinate for a particular value 
of a?i represented the area of the v curve up to that place from some 
datmn value of a?, we could easily solve more difficult problems. I 
have often done this by counting squares on squared paper ; also I 
have worked to obtain a number of points in the new cm-ve by 
using a planimeter a number of times. We see now that if we 
know ii\ the integi-al of iv shows s, the integral of s shows m, the 

integral of — shows i, and the integral of i shows y, the shape of 

the beam. "When we integrate, however, we must settle the start- 
ing value of the new ordinate, and this is what usually gives 
trouble. Thus the starting value of s (when we integrate xc) is 
not zero, but depends upon the supporting forces. 

The student must see very clearly that the change in f, in going 
from one value of x to another, is equal to the area of the m/-e i 
curve between those places. 

359. I have found the arithmetical method of Art. 214 very satis- 
factory. Its accuracy depends, of course, on the number of ordinates 
taken. A student ought to test for himself the accuracy of the 
method on some such exercise as the follo^ving, of which he knows 
the answer. 

A beam of rectangular section \\ inch broad, 2 inches deep, and 
of length 15 inches, is fixed at one end and is loaded uniformly with 
10 lb. per inch. Its e is 25 x 10^, find m everywhere and y. Here 
I = 2-25 X -l^ -j- 12, or I ^ 12. x is distance in inches measui-ed 
from the free end. Imagine the free end to be on our left and the 
rest of the beam on our right. The table gives the whole work, 
and needs almost no explanation. 

The student who thinks will have no difficulty in working out 
a problem of this kind. To find such a number as m for x = 9, for 
example, we add the average value of 130 and 1-40 to the previous 
M, and so get 855. 

We know that i is when x = 15, and so we subtract 

3,752 X 10~ from all the found values of i -t- c In the same 
way to get the last column we add 39,801 to every number of the 
previous column. 

Let us compare our results with the true answers, which the 
student can easily work out as in Art. 339 : s = 50 + 10 .r, m = 50 
a; + 5 a;^ 

^y 1 f „ 5 1 

^ = r^4 "«■«"- "•2«» -+ T ^' + 1 -^ i 

Thus whf n x ==; 15, m = 1,875, as in the table. When x = 0, i or 
di/ldx = — 3,750 X 10~ , whereas the table gives —3,752 x 10~ . 



APPLIED MECHANICS. 



443 



Again, wlienx = 0, «/ = 39,844 x 10 , whereas the table gives 

39,801 X 10^, which is the same for all practical purposes. After 
having worked this example a student must feel confidence in 
using this method of integration which gives us answers so readily. 



X 


w 


s 


M 


M 
£1 


i+ c 


i 

8 


y + c 


y . 










X 10* 


X id 


X 10 


xlO^ 


X 10 





10 


50 










-3,752 





39,801 


1 


10 


60 


55 


18 


9 


-3,743 


-3,747 


36,054 


2 


10 


70 


120 


40 


38 


-3,714 


-7,476 


32,325 


3 


10 


80 


195 


65 


91 


-3,661 


-11,163 


28,638 


4 


10 


90 


280 


93 


170 


-3,582 


-14,785 


25,016 


5 


10 


100 


375 


125 


279 


-3,473 


-18,312 


21,489 


6 


10 


110 


480 


160 


421 


-3,331 


-21,714 


18,087 


7 


10 


120 


595 


198 


600 


-3,152 


-24,956 


14,845 


8 


10 


130 


720 


240 


819 


-2,933 


-27,998 


11,803 


9 


10 


140 


855 


285 


1,082 


-2,670 


-30,800 


9,001 


10 


10 


150 


1,000 


333 


1,391 


-2,361 


-33,315 


6,486 


LI 


10 


160 


1,155 


385 


1,750 


-2,002 


-35,497 


3,304 


12 


10 


170 


1,320 


440 


2,162 


-1,590 


-37,293 


2,508 


13 


10 


180 


1,495 


498 


2,631 


-1,121 


-38,648 


1,153 


14 


10 


190 


1,680 


560 


3,160 


-592 


-39,505 


296 


15 




200 


1,875 


625 


3,753 





-39,801 






360. Beams Fixed at the Ends. — If the loading on a symmetrical 
beam is symmetrical so that we find, graphically or any other way, 
the diagram of bending moment m as if it were supported at the 
ends, we know that equal and opposite torques are needed to fix 
the ends . Thus, if a b is, say, a beam of uniform section and has 
a loading of any kind whatsoever which will produce (the beam 
being only supported at the ends) a diagram of bending moment m 



444 



A.PPLIED MECHANICS. 




Fig. 253. 



such, as is shown iii a c d e b (Fig. 253), and if equal and opposite 
torques are applied to fix the ends such as alone would produce the 
diagram of "bending moment shown to scale in b a f a, then the 

algebraic sum of the two (for 
^ ^ the A c D E B is negative and 

A B G F A positive) is shown 
in A F c' d' e' G B A, being 
positive from a to c' and 
from e' to B, so that in these 
parts the beam is convex 
upwards ; and being nega- 
tive fi'om c' to e', where 
the beam is concave upwards. 
We want, then, to know 
exactly how much g b or 
AF must be to fix the ends of the beam. Now, the difference 
of slope of beam between a and b is nothing if the ends are both 
fixed, and therefore the total area of the bending moment diagxam 
must be zero (since ei is constant, we say m instead of m/ei). 
Hence, the area of the portion c' d' e' being negative, must be 
numerically equal to the sum of the areas a c' f + b e' o. In fact, 
the average ordinate of the 
M curve must be zero, and f 
therefore we raise the m 
cm-ve A c D E b A by its 
average height to get the 
M curve. 

Example 1. — Beam of 
length I, with load av in 
the middle, fixed at the 

ends. The diagTam of m is a d b a (Fig. 254), where d d" re- 
presents 4 w/ ; raise it therefore by the amount | w^, and we find 
A F c' d' e' G B A. Evidently a f = — d" d' ^ g b = -J yfl. The 
beam is convex at the ends, concave in the middle, equally ready 
to break at ends and middle, and the points of inflexion are half- 
way between ends and middle {see Art. 342) 

Example 2. — Beam a b of 
•^ G- length I, with total load w 

spread uniformly. a d b 
shows the m curve, a parabola, 
the diagram of bending mo- 
ment if the beam were merely 
supported at the ends, d d" 
being ^^L The average or- 
dinate of A D B A is I D d". 
Raising the diagram by this 
amount, we find the true 
diagram a r c' d'e' g b a of m for the beam fixed at the ends. There 
is ^^^\l at each end and -_i_ ^yl at the middle (se^ Art. 342). The 
points of inflexion are at c' and e', no longer exactly half-way 
between ends and middle. 

361. Students wiU do well to work at least one complicates 




Fig. 254. 




APPLIED MECHANICS. 



445 



example of a uniform beam fixed at the ends with symm etrical loading. 
If the beam, varies in section, but is symmetrical — that is, if at two 
points equally distant from the two ends the sections are the same, 
and if the loading is symmetrical, first obtain the m curve graphi- 
cally and measure m at a number of equi-distant points. Thus, for 
some beam of 20 feet long, let us suppose that the values of m as 
measured are given in the following table. We need not use b, as 
we will suppose it to be constant. 



Distance 


— m 


Values of 

I 

in inches to tne 

4tli power. 




1 
T 




from end 
in feet. 


in 

ton-feet. 


T 


M. 


1 


1/5 


500 


•03 


•002 


17-85 


3 


Z5 


300 


•0833 


•00333 


7-85 


5 


35 


250 


•140 


•004 


- 2^15 


7 


40 


320 


•125 


•003125 


- 7^15 


9 


44 


360 


•122 


•002778 


-11-15 


11 


44 


360 


•122 


•002778 


-1M5 


13 


40 


320 


•125 


•003125 


- 7^15 


15 


35 


250 


•140 


•004 


- 2^15 


17 


25 


300 


•0833 


•00333 


+ 7-85 


19 


15 


500 


•03 


•002 


+ 17^85 




r 


rotal 


1^0006 


•03046 






1 


Werage 


0^10006 


•003046 





We see that the area of the — curve is to be zero, and if m-^ is 
the unknown bending moment applied at each end to fix it, 

M = m + mj ; so that the average value of ■"• must be zero, 

or the average value of must be equal to m-i multiplied by the 

average value of — . Hence m-i is equal to the average value of 
- divided by the average value of -. 

In the above case this is 0-10006 -^ 003046, so that 
Ml — 32-85 ton-feet; and m = 32^85 + m (algebraical sum) is 
the true bending moment everywhere. 

362. The shearing force diagram in the symmetrical cases is the 
same whether a beam is fixed or onlv supported at the ends ; and 

^^^ ^^ ^°* altered by the fixing, the shearing force and the deflec- 
tion due to shearare everywhere the same in the beam. (See Art. 369.) 



446 



APPLIED MECHANICS. 



The solution just given is applicable to a beam of which the i 
of ever)^ cross-section is settled beforehand in any arbitrary manner, 
so long as i and the loading are symmetrical on the two sides of the 
middle. Let us give to i such a value that the beam shall be of 



uniform strength everj^here ; that is, that - z 



:/.... (2), wher 



2 is the greatest distance of any point in the section from the 
neutral axis on the compression or tension side, and / is the 
constant maximum stress in compression or tension to which the 
material is subjected in every section. Taking z = \d^ where d 

is the depth of the beam, (2) becomes — c? = + 2/ . . . . (3), the 

+ sign being caken over parts of the beam where m is positive, 
the — sign when m is negative. 

As the ai'ea of the - curve fi'om end to end of the beam is 
I 

to be zero, and — :^ 2/,^ we see that the area of a curve show- 
ing everywhere the value of ± \\d ought to be zero, the positive 
sig-n being taken from the ends of the beam to the points of 
inilexion, and the negative sign being taken between the two 
points of inflexion. We see, then, that to satisfy (4) we have 
only to solve the following problem. 

In the figure, eatucgeIs a diagram whose ordi- 

nates represent the values of - or the reciprocal of the depth 

of the beam which may be arbitrarily fixed, care being taken, 

however, that d is the same 
at points w^hich are at the 
same distance from the 
centre, e f g e is a diagram 
of the values of what the 
bending moment m would 
be if the beam were merely 
supported at its ends. We 
are required to find a point 
p such that the area of 
E p T A = area of v o o' t, 
w-here o is in the middle of 
the beam. When found, this point p is a point of inflexion, and 
PR is what we have called m^. That is, m - pr is the real 
negative bending moment m at every place, or the diagram e f g 
must be raised vertically till r is at p to obtain the diagram of m. 
Knowing M and ^, it is easy to find i through (3). 

It is evident that if such a beam of uniform strength is also of 
uniform depth, the points of inflexion are half-way between the 
middle and the fixed ends. 

363. In the most general way of loading, the bending moments 
requii'ed at the ends to fix them are different from one another. 




APPLIED MECHANICS. 



447 



TIius in Fig. 257 let afcgba be Avhat the bending moment m 
would be ii the beam were merely supported at its ends ; let fixing 
moments j«i = a h and ?% =^ e b be applied at the ends, producing 
of themselves {see Art. 354) a bending moment diagram shown by 
A H E B A, or, if A D is a;, then r d, the bending moment produced by 

the end couples is mi + ^ ^ x, iil is the length a b. Let us 

call this Wj + bx. Hence ii or r d - d s is 




Fig. 257 



Now, as I is known, let a Ji J2 J3 J4 b a be drawn to represent 

the value of — everywhere. Let it be integTated ; that is, let such 

an ordinate as d k^ (call it y) represent the area of a j^ j^ d, and so 
obtain a Ki Kg K3 b a, and let Yi be the value of this when x = ab. 

Also integrate - , and say that the ordinate of the resulting curve 




A Li Lg Lg B A is X, and let Xj be the value of this when x = ab. 
Tnteerate also the curve whose ordinate is — everywhere, calling 



448 APPLIED MECHANICS. 

the answer fi, and /i^ its value for a; = a b ; then (2) becomes, 

since ~ is the same at both ends, 
dx 

= ;;?j Tj + 5 Xi - ,Uj . . . . (3). 

Again integrating t, x. and ju (but it is only necessary to get the 
areas over the whole length at once), calling the answers Yj, Xj, 
and Mj, we see that, since at the two ends y is 0, 

= m^Y^ + iXj - Ml . . . . (4). 

The two unknowns, m^ and b, can now be foimd from (3) and (4).* 
We give in Art. 365 an example completely worked out. The 
column headed fJi represents the bending moment in ton-feet due 
to a given set of loads, if the beam were merely supported at its 
ends. These values may be found, of course, by the graphical 
method of Art. 349. The values of i are supposed to be given us, 
and they are in inches to the fourth power. 

364. "We shall now consider a beam fixed at one end, b, and 
merely supported at the other, a, which is on the same level as b. If, 
as before, m is the bending moment at any place, d, which would 
exist if the beam were supported at each end, and if ^w^ is the 
fixing couple, the true bending moment is 

Take, first, a simple case, a uniform beam uniformly loaded with 
w lb. per inch. It is easy to prove, as in Art. 339, that w = 
- ^wl.v + I icx-, and 

•£.!-—= -Y^ - \ ^v^x + \ icx^ .... (2). 
Hence 

^i% = f^x'^ - ItvJx'^ + \u:^ + c . . . . {^), 

* We have used the symbols M, X, T, /^i, Xi, Ti, M, x, Y, Mi, Xi, Yj, fearing 
that students are still a little unfamihar with the symbols of the calculus. 
Perhaps it would have been better to put the investigation in its proper 
form, and asked the student to make himself familiar with the usual symbol, 
instead of dragging in fresh symbols. 

After (3) above, write as follows :— 



Elizl =mA ^ + M ^^^-^-1 ^cfx = 0....(4). 



l-Jq Jo Jo •-'o 




Again integrating between limi 

I 



D 




^M 1 T"^n I ~1 1 , j----0....(5). 

J 0^ 



The integrations indicated in (4) and (5) being performed, the unknowns m\ 
and h can be calculated and used in (1). The student must settle for himself 
which is the better course to take— to use the formidable-looking but really 
easily understood symbols of this rxote, or to introduce the letters whose 
meaning one is always forgetting. 



APPLIED MECHANICS. 449 



and 

Elf/ =Q^f^ - T2 «"^^^ + 
aV u-x^ + cx . . . . (4). 

We need not add a 
constant here, because y 
is when .r is 0. In 
(3) and (4) we insert the 
conditions that 



■ = when x = I 
dx 

y = when x = I , 


(6), 


and we find 




C 







" 6 



Fig. 259. 
2V loV^ + 'J (7). 



IFrom these we find that c = Jg ivP, w^ = | wlr. so that the true 
bending moment everywhere is given by (2), and the shape of the 
beam by (4). The slope of the beam at the su^iported place is 

W^3/48 E I. 

If the loading is of any kind whatsoe'S'er, and if the section 
varies in any way, it is necessary to be able to integrate a function 
when given as the ordinate of a curve. We have, as before, in ( 1 

Ef^ = ?l^ £+'«.... (8). 
dx^ C 1 1 ^ ' 

Let the integral of - be called x, and the integral of be called ^u, 

and let curves be drawn showing the values of x and fx. Let their 
values when a; = ^ be Xj and ^uj. Then 

4 = ^x. ...... ..(9). 

We can now integrate (9) again, and obtain e?/; but if we do 
jDot actually wish to find the shape of the beam, we need only use 
ths two ideas — first, that ?/ = when re = 0, and this shows that no 
new constant needs to be added ; secondly, that y = when x = I; 
and hence, if the areas of the whole x and /j. curves from a; = to 
a; = ^ are Xj and Mj, we have 

0= Y^X; +Mi + C^ (10). 



450 APPLIED MECHANICS. 



Also, using in (9) the statement that -^ is zero when x = /, wo 



have 



dx 



= ^^X,-;x, -f c....(ll). 



From (10) and (11) we can find c and — ^. Using the value of m^ 

BO found, we find the bending moment everywhere given in (1).* 

If in the above exercise we imagine the supported end at a to 
he raised a distance a above its original position, or that b has 
settled downwards by this amount, find (supposing the beam, fixed 
at one end, b, to be unloaded) what upward force at a Ccall it p) 
will cause it to rise through the distance a. We have only to 
assume that there is an additional supporting force of thb amount, 
and that the bending moment due to it acts as well as the other- 
in feict, instead of (1), we have as the bending moment 



dx- I 

365. It is very important that the student should work out care 
fully such an exercise as the following. A beam is given with loads, 
and we know m and i at every place. The integrations to be 
performed are much the same whether it is a case of^ a beam fixed 
at the ends or fixed at one end and merely supported at the other, 
and therefore we give both. The two results are stated imme- 
diately after the table. 

* "Withont using the letters ,u, x, .Uj, Xi, etc., the above investigation is 

Integrating (9) between the limits and /, and recollecting that y is zero at 
ooth Uiiaits, 

= X \.A ' = f f IV- + r ff '>- + rf • • • • m. 

Also using in (9) the statement that -'7 is zero when x = ?, we have 

I 

.(11). 



o-H =?Ji-+j>-^ 



The integrations in (10) and (11) being performed, the imknowns j and e 
can be calculated. The true bending moment ererjwhere is what we started 



with, wi + -T-«. 



APPLIED MECHANICS. 



451 











Y = 




X = 




jU, = 




M 


X 


m 


I 


1 

I 


fid. 

I 


X 

T 


Ji- 


7)1 

r 


I-'- 


M 

Beam 

Fixed at 

Eucls. 


Beam Fixed 
at one End, 
Supported at 










^ 




%J 




»^ 


the Other. 




~ 


+ 


+ 


+ 


+ 


+ 


~ 


— 














•0 














'5 


8 


5^0 


•00200 


•00200 


•00100 


•00100 


•01600 


•01600 


+ 22^30 


- 7-13 


1-5 


15 


450 


•00222 


•00422 


•00333 


•00433 


•03330 


•04930 


+ 14-73 


- 12^38 


2-5 


21 


400 


•00250 


•00672 


•00625 


•01058 


•05250 


•10180 


+ 8^16 


- 16-64 


3-5 


25 


350 


•00286 


•00958 


•01001 


•02059 


•07150 


•1733 


+ 3^59 


- 18-90 


4-5 


28 


320 


•00313 


•01271 


•01404 


•03467 


•08740 


•2607 


+ ^03 


- 20^15 


5-5 


32|300 


•00333 




•01826 




•1066 




- 4^54 


- 22^41 










•01604 




•05289 




•3673 






6-5 


34 


300 


•00333 


•01937 


•02159 


•07448 


•1132 


•4805 


- 7-11 


- 22^66 


7-5 


36 


320 


•00313 


•02250 


•02340 


•09788 


•1123 


•5928 


- 9-67 


- 22^92 


8-5 


37 


350 


•00286 




•02431 




•1058 




- 11-24 


- 22^18 










•02536 


•1222 




•6986 






9-5 


35;400 


•00250 


•02375 




•0875 




- 9-81 


- 18^43 










•02786 




•1459 




•7861 






10-5 


32 


400 


•00250 


•03036 


•02625 


•1722 


•0800 


•8661 


— 7^38 


- 13^69 


11-5 


31 


400 


•00250 


•03286 


•02875 


•2009 


•0775 


•9436 


- 6^95 


- 10^94 


12-5 


30 


380 


•00263 


•03549 


•03288 


•2338 


•0789 


r0225 


- 6-51 


- 8^20 


13-5 


28 


360 


•00278 


•03827 


•03753 


•2713 


•0774 


r0999 


- 5^08 


- 6^46 


14-5 


26 


330 


•00303 


•04130 


•04393 


•3153 


•0788 


r7187 


- 3^65 


- -71 


15-5 


24 


300 


•00333 


•04463 


•05161 


•3669 


•0799 


1-25S6 


- 2^22 


+ 3^03 


16-5 


18 


330 


•00303 


■04766 


•05000 


! 

•4169 


•0545 


r3i3i 


+ 3^22 


+ 10^78 


17-5 


12 


380 


•00263 


■05029 


•04602 


•4629 


•0316 


1^3447 


+ 8^65 


+ 18^52 


18-5 


5 


400 


•00250 


•05279 


•04625 


•5092 


•0125 


1^3572 


+ 15^08 


+ 27-26 


19 5 


4 


500 


•00200 




•03900 




•0080 




+ 15-52 


+ 30.01 








Yl = 


•05479 


Xi = 


•5482 


M, = 


r3652 








•5748 


4^062 


15-276 















452 APPLIED MECHANICS. 

Beam Fixed at the Ends. — ^i + m^ Yj + p x^ = 0. where wi 13 

the bending moment where x = 0, m^ where z = I, t = ^-^— — -^, 

or - 1-365 + -0548 ;«i + '548 p = ; also - Mj + /«j yi + p Xi = 0, 
or - 15-276 + -5748 m^ + 4-062 p = 0. Hence, ;«, = 30o8, 
p = - -5673, 1)12 = 19-23. Adding p a; to ni, we find the true 
"bending moment m given in the tenth column of the table. It will 
be found that if every value of m be divided by the corresponding 
value of I, the algebraic simi for the whole beam is - -0013 
instead of 0— a very close approximation to accuracy. The 
student may easily proceed to find the shape of the beam. 

Beam Fixed at One End.— Where x = I and the bending 
moment is m^. supported merely at the other, where a; = 0, 

- /^3 + PSi + (? = 0, 

where c is e V at a; = 0. or - 1-365 + -5482 p + c = 0: 
dx 

- M, + p . Xi + f? = 0, 

or - 15-276 + 4 062 p + 20 f = 0. 

Hence p = y = l"^^-- "^^ '"'2 = ^^'SS, and c = -4091. Adding 

p a; to m everywhere, we find the true bending moment m given in 
the eleventh column of the table. The student may easily proceed 
to find the shape of the beam. 

366. As we have proved (Art. 357), since 

du . di M ^M ^ ^s 

-f = I, — = —, — = S, ;t- = w, 

dx dx E I dx dx 

we have a succession of curves which may be obtained from 
knowing the shape of the beam y by differentiation, or which may 
be obtained from knowing ic, the loading of the beam, by integra- 
tion. Knowing ic, there is an easy graphical rule for finding m ; 

knowing — , we have the same graphical rule for finding y. 

Some rules that are ob-sdously true in the tv to ii construction and 
need no mathematical proof may at once be used without mathe- 
matical proof, in applying the analogous rule from — to y. Thus 

the area of the — cm^ve between the ordinates Xi and Xo is the 

EI '^ 

increase of t from Xj to x^, and tangents to the curve showing the 
shape of the beam at ar^ and x^ meet at a point which is vertically 
in a line with the centre of gravity of the portion of area of 

the curve in question. The whole area of the — curve in a 

span H J is equal to the increase in -^ fi'om one end of the span to 

the other, and the tangents to the beam at its ends h j meet in a 



APPLIED MECHANICS. 453 

point p, wtLich is in the same vertical as the centre of gravity of the 
whole — curve. These two riiles may be taken as the starting- 
point for a treatment of the most difficult problems in beams 
by graphical methods. 

If the vertical from this centre of gxa^dty is at the horizontal 
distance h g from h and g j from j, then p is higher than h 
by the amount h g x i j^, the symbol i ^ being used to mean 
the slope at h ; j is higher than p by the amoimt gj x i at j. 
Hence j is higher than h by the amount uq . i^ + Gj.tj 
a relation which may be iiseful when conditions as to the relative 
heights of the supports are given, as in continuous beam problems. 

367. Theorem of Three Moments. — For some time railway engi- 
neers, instead of using separate girders for the spans of a bridge, 
fastened together contiguous ends to prevent their tilting up, and 
so made use of what are called continuous girders. It is easy to 
show that if we can be absolutely certain of the positions of 
the points of supjDort, continuous girders are much cheaper than 
separate girders. Unfortunately a comj)aratively small settlement 
of one of the supports alters completely the condition of things. 
In many other parts of applied mechanics we have the same 
difficulty in deciding between cheapness with some uncertainty 
and a greater expense with certainty. Thus there is much gxeater 
uncertainty as to the nature of the forces acting at riveted joints 
than at hinged joints, and therefore a structure with hinged joints 
is preferred to the other, although, if we could be absolutely 
certain of our conditions, an equally strong riveted structure 
might be made which would be much cheaj)er. 

Students interested in the theory of continuous gii'ders will do 
well to read a paper published in the '• Proceedings of the Royal 
Society," cxcix., 1879, where they will find a graphical method of 
solving the most general problems. We shall take here, as a good 
example of the use of the calculus, a uniform girder resting on 
supports at the same level, with a uniform load distribution 
on each span. Let abc be the centre line of two spans, the 
girder originally straight, supported at a, b, and c. The distance 
from A to B is I^ and from b to c is /g, find there are any kinds of 
loading in the two spans. Let a, b, and c be the bending moments 
at A, B, and c respectively, coimted positive if the beam is convex 
upwards. 

At the section at p at the distance x from a let m be what the 
bending moment would have been if the girder on each span were 
quite separate from the rest. We have already seen that by 
introducing couples m.2 and m^ at a and b (tending to make the 
beam convex upwards at a and b) we made the bending moment at 
p really become what is given in Art. 363. Our m^ = A, wj = b, 
and hence the bending moment at p is 

B - A drp .,. 



454 



APPLIED MECHANICS. 

where m would be the "bending moment if the beam were merely 
supported at the ends ; and the supporting force at a is lessened by 
the amotmt 



A - 



^....(2). 



'1 

Assume e i constant and integrate with regard to x, and we have 

It 1 o b - a du ,„, 

\rn.dx-\-AX + ^x^ — ; — -f ffj = E I . J- . . . . (3). 



TTsing the sign l | >n dx . dx to mean the integi-ation of th€ 
curve representing I m . dx. we have 



fl 



m .dx .dx + ^Ax- + ^ r^ — \- c^x -i- e = t.1 .y . . . . (i). 

As y is when x = and it is evident that I \m.dx.dx = 
when X = 0, e is 0. Again, y = when x = ^i. ITsing the 
symbol fix to indicate the sum I \ m . dx . dx over the whole span, 



Si' 



t^i + i^¥ + i^iH^ - ^) + cih = 0.... (5). 

From ''3 let us calculate the value of e i -f at the point b. and 

dx ^ ' 

let us use the letter a^ to mean the area of the m curve over the 



Jo 



span, or I >'i . dx\ so that e i ^ at b is 

ffl + A 7^ + i /i (b - a) + Cj (6). 

But at any point q of the second span, if we had let b Q = x, we 
should have had the same equations as (l^, (3), and (4). using the 
letters b for a and c for b and the constant C2. Hence, making 

this change in (3), and finding ei ^ at the point b where a: = 0, 

we have (6; equal to c-i or 

f2 - Ci = fl^ + A fj + ^ 7i (B - a) (7), 

and instead of (5) we have 

^2 + \Y,l^ + W (c - B) + r,', = . . . . (8). 
Subtracting (5) from (8), after diA-iding by \ and /gj ^e have 

r3-., = ^-^' + iA?i-lB/2 + i/i (B-A)-|/2 (C - B) .... (9). 
'1 '2 



APPLIED MECHANICS. 455 

The equality of (7) and (9) is 

A/j + 2b (^j + ^2) + c/2 = 6 (^1 _ aj - |2^ . . . . (10), 

an equation connecting a, b, and c, the bending moments at three 
consecutive supports. If we have any number of supports, and at 
the end ones we have the bending moments because the girder is 
merely supported there, or if we have two conditions given which 
will enable us to find them in case the girder is fixed or partly 
fixed, note that by writing down (10) for every three consecutive 
supports we have a sufiicient nimiber of equations to determine all 
the bending moments at the supports. 

Example. —Let the loads be u\ and zro per imit length over two 
^ jonsecutive spans of lengths \ and l^. Then 

?» = - i iclx + h. wx^, \m . dx = - ^ tvlx^ + ^ ivx^, 






ind '^i = - fo ^1^' I \»^ ■ d^ • <^^ = - t\ ^^^^'^ + -h ^^^» 

and )«i = - ^"if^ihS H= - -h'^'zh*- 

Hence ^' + a-^ - j- becomes - Jj trs^a^ - ^-^ ^^^ ^ ^ ^^^^3^ 

or - ^ [w^h]^ + Wil^), 

and hence the theorem becomes in this case 

A^i + 2 B {h + h) + Gh = I {^'2h^ + Wi?!^) no). 

If the spans are similar and similarly loaded, then 
A + 4b + c = iw;?2 (11). 

Case 1. — A uniform and uniformly loaded beam rests on 
three equidistant supports. Here a = c = and b = + ^wt^. 
m = - ^to {Ix - x'^), and hence the bending moment at a point 

p distant x from a is - ^w {Ix - x^) + - -. ^ wP. The sup- 
porting force at A is lessened from what it would be if the part of 
the beam ab were distinct by the amount shown in (2) — - — 

or I wl. It would have been ^ wl, so now it is reaUy | wl at each 
of the end supports, and as the total load is 2 wi, there remains 
^wl for the middle support. (See also Example on next page.) 

Case 2. — A uniform and uniformly loaded beam rests on four 
equidistant supports, and the bending moments at these supports 
arc A, B, c, D. iSTow, a = d =: 0, and from symmetrv b = c. 
Thus (11) gives us + 5b = Jw--^^ or b = c = -^-^lul^. If 
the span a b had been distinct, the first support would have had 
the load ^wl; it now has ^icl - ^^wl or ^^tvl. The supporting 
force at d is also j% id. The other two supports divide between 
them the remainder of the total load, which is altogether 3 wl, and 



456 



APPLIED MECHANICS. 



The supporting forces are then 



so each receives ii tcl. 
ii wl, ii lul, and ^ wl. 
Exercise. — If a heam a b c has any kind of loading 



io 



wl. 



and varies in section 
in any way, and is supported at three places, a and c, on the same level, b, 
on a level b inches below a or c, first find the diagram of bending 




Fig. 



moment and the deflection y^ everj^where of the beam, assuming the 
s^ipport B not to exist. Find, in particular, y^, the deflection at the point 
B. Kow consider a new problem — the same beam supported at two places, 
and with only an upward force at b. Find what the force b must be to 
cause a deflection upwards, yi - b, at k, and what the upward deflection z 
everywhere is. This force b is evidently the supporting force at b in the 
real problem, and the deflection in the real problem everywhere is y - z. 

Example. — Uniform loading w per inch on a b and on b c, each 
of length I ; beam of uniform section, the supports all on same 
level. (1) If prop b is absent, the deflection at b would be (Art. 

339) ^ ^^^^- ^^^^l CaU this S. Each of the supportinc 



forces is 



wl. 



384 EI 

(2) Beam of length 27, supported at the ends. If an upward 



force B produces an upward deflection, we know that 5 = Jg- — {2l)K 

-nr ^ , B(2 03 5 2wl . (2^3 . , , . . 

Hence we have ^ = -ttttt , and b = ^tv? — that is, 

*^ EI 384 EI ' * ' 

when a uniform and uniformly loaded beam is on thi^ee equidistant 
supports, the supporting forces axe fwl, ^yivI, and ^wl. 

Mr. George Wilson (Proc. Eoyal Soc, Nov., 1897) describes a 
method of solving the most general problems in continuous beams 
which is simpler than any other. Let there be supports at abode. 
(1) Imagine no supports except at a and e, and find the deflections at 
B c and D. Jn'ow assume only an upward load of any amount at b, and 
find upward deflections at b c and d. Do the same for c and d. 
These equations enable us to calculate required upward loads at b o 
and D, which will just bring these points to their proper levels. 



467 
CHAPTER XXI. 

BENDING AND CRUSHING. 

\. Stress over a Section. — When any portion of a column 
or beam or arch on one side of a section, b c, is acted upon by 
loads and supporting forces, we can generally find one force, 
representing the resaltant of the stresses at the section, which 
will balance them all. If, instead of a force, we merely get a 
couple, then the section is exposed solely to bending moment, 
and we know now how to find the effect of this. If the force 
is parallel to the section, then we know that the section is 
either exposed to mere shearing strain or shearing and bend- 
ing, as in a horizontal beam with vertical loads ; but if the 
force is inclined to the section, there will usually be shearing 
and bending, and besides this a uniform distribution of com- 
pression or extension all over the section. In practice we 
generally find that compression and bending alone have to be 
considered. Thus, if bc (Fig. 261) is the 
edge view of the section of a structure, 
being a line at right angles to the paper 
through its centre of gravity, and if f is 
the resultant of all the forces supposed in 
the plane of the paper which act on the 
structure to the right of the section, let 
p and s be the resolved parts of P normal 
to and tangential to the section ; then s is 
balanced by an equal and opposite shear- 
ing force which must be exerted by the Fig. 26i. 
material to the left, p is a compressive 
load which is spread uniformly over the section, producing 
a compressive stress p/a if a is its area ; but besides this we 
have in the section the varying compressive stress on the b 
side of o and the tensile stress on the c side of o, which the 
bending moment p. od produces. In fact, the compressive 
stress at any place which is at the distance y from o on the 

. . . p p . o D ^^. 

compression side is - + y . . . , [1). 

The student ought to draw a slice between two parallel 
cross-sections like b c near one another, and draw the change 




458 APPLIED MECHANICS. 

of shape, first making it uniformly thinner because of p/a, and 
then making it wedge-shaped. We have, in fact, the wedge- 
shape G e f d' of Fig. 207, where h'j' = hj, and in addition 
we have to imagine e f moved parallel to itself in the direc- 
tion c d' ; but students must draw this for themselves. They 
will see that the result may be — compression everywhere, but 
much more at b than at c ; or compression everywhere and 
just no stress at c ; or compression at b, tension at c, and 
a neutral line somewhere between. Most people have the 
habit of calling the line through o the neutral line of the 
section, although it is the neutral line only when the com- 
ponent p is zero. 

The proof of the ahove statement is this : Assume that a plane 
section remains plane, it follows, as it did in Art. 319, that there is 
a neutral line, say at o', at right angles to the paper. Let any 
point H be at the distance ~ from o' ; the compressive s ress there 
is cz say, where c is some constant. The force on a small area a 
there, is cza. Then p = c 2 z« . . . (2) and t . OJi = c^za .y . . . (3), 
because, if o h = ?/, za .y is the moment of za about o. Now 
z = y + o o', so that equation (2) is 

p = c.oo'2a + c2y.a = c.oo'.A.... (4), 

because ^ya = o ; and p . o d = c 2 y'^a, + e'2,oo'.y.a = ci.... (5) 
if I is the moment of inertia of the area about o. We see that 

POD P 

cy from (5) is ~ y and c . o o' from (4) is - ; but the sum of 

these is C2, the compressive stress, and so we have proved (1). 

If the section is rectangular, the dimension at right angles to 
the paper being 1, and b c being d, then i = 1 x d^/12 and 
A = 1 X ^. The compressive stress is least at c where y = - \d, 

P 6 P . O D 

or by (1), the compressive stress at c is ^ - — -^ — .... (6). 

As o D gets greater, the compressive stress at c becomes less until 
there is a value of o d w^hich just causes c to have no stress ; 
a greater value of o d than this would create tensile stress at c. 
We usually take it that in a masonry joint there ought to be 
no tensile stress, and hence in a masonrj- joint the hmiting value 
of D is given by putting (6) equal to zero ; that is, 6 . o d = <f. 

Hence D must fall within the middle third of the masonry 
joint, B c, if there is to be stability. This is the fundamental 
rule in the design of arches and buttresses. Another condition 
of stability for a masonry joint is that s shall not exceed the 
frictional resistance, or the angle J D L must be less than the 
angle of repose (see Art. 96) for the materials. 

The above rule is very generally useful in machine design, 



APPLIED MECHANICS. 



459 



but we need not give many examples. Fig. 262 is a crane-hook. 
The section at bc is not usually elliptic ; rather like an ellipse, 
with the end at B blunter than that 
at c. If is a line at right angles to 
the paper through the centre of grav- 
ity of the section, and w is the load, 
the stress at any place is that due to 
a bending moment w . o D, together 
with a tensile stress due to w being 
spread uniformly over the section. 

When a weight w hangs from a 
bracket as in Fig. 263 the strength 
at any section, such as o, is merely 
calculated from the bending moment 
w . D B, because when the distance 
D B is con- 
siderable the 
stresses due 
to the bend- 
ing are usu- 
ally much 
greater than 
those due to 

w, divided by the area of cross- section. 
369. Shear Stress in Beams. — Let the 
distance measured from any section of a 
beam, say at o (Fig-. 264), to the section 
at A he x^ and let OB = a;-t-Sa;. Let the 
tending moment at c' a c he m, and at 
b' B D be M + Sm. Let a c be the com- 
pressive side of c c'. Let o ab (Fig. 264) and a a (Fig. 265) repre- 
sent the neutral surface. We want to know the tangential or shear 
stress /s at e on the plane c a c'. Now it is known that this is the same 





Fig. 263. 





c D 






0— 


A 




n 














E 




F 








H 











Fig. 264. 




/I ^ ^ 

H M 



C D 

Fig. 266. 



as the tangential stress in the direction e p on the plane e p, which 
is at right angles to the paper, and parallel to the neutral surface 
at a b. Consider the equilibrium of the piece of beam e c d p, shown 



/..ar.>^«d] 



i ETiti BEESSEIir It 



-^^L-'-m/\-'^ 












APPLIED MECHANICS. 461 

we see that at any point of this heam x inches, measured 
horizontally from the middle, and y inches from the neutral axis, 
the shear stress is 

Observe that where y = the shear stress is greater than 
at any other point of the section — that is, at points in the 
neutral axis. The shear stress is zero at c. Again, the end 
sections of the heam have greatest shear. A student has much 
food for thought in this result (5). It is interesting to find the 
directions and amounts of the principal stresses at every point of 
the heam — that is, the interfaces at right angles to one another at 
any point across which there is only compression or only tension 
without tangential stress. 

We have been considering a rectangular section. The student 
ought to work exercises on other sections as soon as he is able to 
integrate by with regard to y in (1) where h is any funciion of y. 

J AC 
hy . dy is equal to the area of e h c h e 
AE 



(Fig. 265) multiplied hy the distance of its centre of gra^dty from 
A A. Taking a flanged section, the student will find that/, is small 
in the flanges, and gets greater in the web. Even in a rectangular 
section /g became rapidly smaller further out from the neutral axis ; 
but now to obtain it we must divide by the breadth of the section, 
and this breadth is comparatively so great in the flanges that there 
is practically no shearing, the shear being confined to the web; 
whereas in the web itself / does not vary verj'- much. The student 
already knows that it is our usual custom to calculate the areas of 
the flanges, or top and bottom booms of a girder, as if they merely 
resisted compressive and tensile forces, and the web or the diagonal 
bracing as if it merely resisted shearing. He will note that the 

shear in a section is great only where -=-, or rather -^- ( — V is 

dx dx \ij' 

great. But, inasmuch as in Art. 357 we saw that -^ = s, the 

total shearing force at the section, there is nothing very extra- 
ordinary in finding that the actual shear stress anjnvhere in the 
section depends upon its value. 

Defllection of Beams due to Shear.— If a bending moment 
M acts at a section of a beam, the part of length ?x gets the 
strain-energy ^ m^ 8x/e. i, because m dxj-E i is the angular change, 
and therefore the whole strain-energy in a beam due to bending 
moment is 

2 

dx (6). 



2eJ I 



2 

If / is a shear stress, the shear strain-energy per unit volume 



462 



APPLIED MECHANICS. 



is /2/2n .... (7), and by adding we can therefore find its total 
amount for the whole beam. 

By equating the strain-energy to the loads multiplied by half 
the displacements produced by them we obtain interesting rela- 
tions. Thus in the case of a beam of length I, of rectangular 
section, fixed at one end and loaded at the other with a load w ; 
at the distance x from the end, m = wjt, and the energy due 
to bending is 

4 



j_rw^ 

2e I I ■ 

•^ 



dx=yf'^PleBi (8). 



The above expression (5) gives for the shearing stress 



1 6 



/=j^3a-^-y')^^----(9)- 



The shear strain-energy in the elementary volume b. Sx. 5y is 
b. dx. 5y./2/2N. Integrating this with regard to j/ from - ^6? to 
+ ^d, we find the energy in the slice between two sections to be 
3 w2/. Sx/o N b d, so that the shear strain-energy in the beam is 
3 w2Z/5 N * ^ . . . . (10). 

If now the load w produces the deflection z at the end of the 
beam, the work done is ^^z . . . . (11). Equating (11) to the 
sum of (8) and (10) we find 



3ei "^ 5 



wZ 
^bd' 



.(12). 



Note that the first part of this due to bending is the deflection 
as calculated in Art. 339, Example 1. ^Ye believe that the other 
part due to shearing has never before been calculated. 

370. Springs which Bend. — We consider the bending in springs 
of regular shape, such as spiral springs, later, in Art. 521. But it 

seems natural to 
consider certain 
irregularly shaped 
springs here. Let 
Fig. 267 show- the 
centre line of a 
sj)ring Jixed at a, 
loaded at b with 
a small load w 
in the direction 
show-n. To find 
the amount of 
yielding at b, the load and the deflection are supposed to be very 
email. Consider the piece of spring bounded by cross-sections at 
r and q. Let p q, = 5s, the length of the spring between b and 
p being called a. 




Fig. 267. 



APPLIED MECHANICS. 463 

The bending moment at p is w . p r ot w . x, ii x is the length 
of the perpendicular from p upon the direction of w. Let br be 
called y. Consider fii'st that part of the motion of b which is due 
to the change of shape of qp alone; that is, imagine aq, to be 
perfectly rigid and p b a rigid pointer. The section at q being 
fixed, the section at p gets an angular change equal to 5s x the 

change of curvature there, or 5s — or — '- .... (1), where 

° EI EI ^ ' 

E is Young's modulus and i is the moment of inertia of the cross- 
section. The motion of b due to this is just the same as if p b 
were a straight pointer ; in fact, the pointer p b gets this angular 
motion, and the motion of b is this angle multiplied by the straight 

distance p b, or — '- . p b . . . . (2). Now how much of b's 

EI ^ ' 

PR 

motion is in the direction of w ? It is its whole motion x — 

P B 

or X — , and hence b's motion in the direction of w is — ^ . . . (3). 

p B E I ^ ^ 

Similarly b's motion at right angles to the direction of w is 

In the most general cases it is easy to work out the integrals of 
(3) and (4) niunerically. We usually divide the whole length of 
the spring from b to a into a large mmiber of equal parts so as to 
have all the values of 5s the same, and then we may say (s being 

the whole length of the spring) that we have to multiply --^- 
upon the average values of — and — for each part. In a well- 
made spring, if b is the breadth of a strip at right angles to the 
paper and t its thickness, so that i = -^ bfi, we usually have the 

spring equally ready to break everywhere, or „ = /> a constant. 
When this is the case (3) and (4) become -^ and 

E t 



—-^— . -. And if the strip is constant in thickness, varying in. 

breadth in proportion to x, then • ■ . x is (3) and -^ — . y 

is (4). If :^and y are the x and y of the centre of gravity of the 

curve (see Art. 110), -^^ is the total yielding parallel to w, and 

thig is what we generally desire to know. ^ ' ^ is the total 
yielding at right angles to w. 

371. Struts. — As we have already said, when a very short 
column is loaded (as it usually is) in such a way that the 
material is prevented from swelling laterally, it will withstand 
exceedingly great loads without fracture. It is only when thQ 



464 APPLIED MECHANICS. 

length of a square or round prism is three or four times itg 
diameter that the material gets a chance of showing how it 
behaves under compressive stress ; up to a length of about ten 
times its diameter its breaking load is now nearly the same, and 
it remains the same for any length if slight lateral restraints 
against bending are provided. In a tie-rod, want of uniformity 
of stress causes yielding, and this is the same in a strut ; but 
in a tie the effect of yielding is to remedy the defect, the tie 
gets straighter, the stress better distributed, whereas in a strut 
local failure causes bending, instability, and a worse distribu- 
tion of stress. In Art. 243 we have discussed the behaviour 
of material under uniform axial compressive stress merely, and 
also in columns which are too short for mere axial compressive 
stress. We now know that if b c. Fig. 261, is a cross section of 
a strut, and if the resultant load on the part of the strut on 
one side of b c is F or p, not only have we the usually assumed 
stress p/a over the section, but also the bending 
moment stress. What this may amount to de- 
pends upon D ; and this depends on two things 
— first, the want of accuracy in applying the load 
at the ends of the strut, a matter which cannot be 
taken into account in calculation unless we know 
how much error there is ; secondly, the bending 
of the strut. 

372. Bending of Struts. — Consider a strut per- 
fectly prismatic, of homogeneous material, its own 
weight neglected, the resultant force F at each end 
passing through the centre of each end. Let A c B, 
Fig. 268, show the centre line of the bent strut. Let 
p Q = 2/ be the deflection at p where oq=zx. Let o A 
z= OB = I ; yis supposed every wli ere small in com- 
parison with the length 21 of the strut. Let e be 
Young's modulus for the material, and i the least 
moment of inertia of the cross-section everywhere, 
j^,.^^ ^^g about a line through the centre of gravity of the 

section. Then the result of the following theory is 
that the load f, which will produce bending, is e i tt'^JU-. 

F y 
r y is the Dendine- moment at p and — is the curvature there. 
-^ ^ EI 

Then as in Art. 339 the curvature being — -4 we have 

TO ^^^ 

EI dx^ ' ' ' ' ^ ^' 



APPLIED MECHANICS. 465 

Notice that when we choose to call ^-^ the curvature of a curve, 
if the expression to which we put it equal is essentially positive, we 
must give such a sign to -j-^ as will make it also positive. Now if 
the slope of the curve of Fig, 268 be studied as we studied the 
curve of Art. 338, we shall find that -^-^ is negative from a; = to 

a? = o A, and as y is positive so that — ^ is positive, we must use 

£ I 

— 3-^ on the right-hand side. 

If the student tries he will find that 

y = a cos. X A /— .... (2) 

satisfies (1) whatever value a may have. When a; = we see that 
«/ = <?, so that the meaning of a is known to us ; it is the deflection 
of the strut in the middle. The student is instructed to follow 
carefully the next step in our argument. When x = l^ y = 0. 
Hence 



« cos. I 



V E I 



• (3;. 



How can this be true ? Either a = 0, or the cosine is 0. 
Hence, if bending occurs, so that a has some value, the cosine must be 

0. Now if the cosine of an angle is 0, the angle must be - or — 

or -^, etc. If we confine our attention to -x, the condition that 
2 2i 

"bending occurs is 

is the load which will produce bending. This is called Euler's law 

of strength. It is easy to see why we confine our attention to - 

as it gives the least value of w. The meaning of the other cases is 
that y is assumed to be one or more times between x =. ^ and 
iP = ^, so that the strut has points of inflexion. 

It will be found that the complete solution of any such equation 
dP-ii 
as (1), which may be written ^^-?^2y=0, \% y—a cos. nx->rh sin. nx 

where a and h are arbitrary constants, a and b are chosen to suit 
the particular problem which is being solved. In the present case 
it is evident that as y = when x =■ I and also when x — - I 
= a COS. nl -^^ b sin. w^, = a cos. nl - h sin. nl, so that b is 0. 



466 APPLIED MECHANICS. 

Now let -us consider a strut fixed at the ends. The resultant 
force F at an end no longer acts through the centre of gravity of 
the end section. The solution is as hefore 

y = Cb cos. nx + h sin. nx . . . . (5). 
Differentiating we have 

-^ = - na sin. nx + nb cos. nx . . . . (6). 

Now ^ = for a: = 0, a; = ?, a; = - l^ and hence = nh^ so that 

^ = or ^ =■ a COS. nx . . . . (7). 

Also = - na sin. nl = na sin. nl (8). 

Now (3) tells us that a is the deflection at c, where x is 0, and if a 
has any value, (4) tells us that sin. nl is — that is, nl = tt, or 



^f/e I = tt, or r = -p- .... (9). 



Hence if a strut is fixed at both ends the load which it will 
stand before bending is the same as for a strut of half the length 
hinged at the ends. In fact, if g c H is a strut fixed at the ends, 
its strength is like that of a strut k l hinged at 
K and L, or again like that of a strut G K l fixed at G 
and hinged at l. The strength to resist bending of 
a strut of length 2 I hinged at the ends may be cal- 
culated from (4) ; this is the strength of a strut of 
length 3 Z, hinged at one end and fixed at the 
other, or of a strut of length 4 1 fixed at both ends. 
|Q. It is only necessary, then, to remember the rules for 
struts hinged at the ends. 
^' The load given by (4) will produce either very 

little or very much bending equally well, and we 
may take it that p given by (4) is the load which 
will break a strut if it breaks by bending. If /is 
the compressive stress which will produce rupture 
and A is the area of cross section, the load /a will 
break the strut by direct crushing, and we must take 
the smaller of the two answers. In fact, we see 
Fio-. 265) tha^t f K is to be taken for short struts or for struts 
which are artificially protected from bending, and 
(4) is to be taken for long struts. Now, even when great care is 
taken we find that struts are neither quite straight nor homo- 
geneous, nor is it easy to load them in the specified manner. 
Consequently, when loaded they defiect with even small loads, 



APPLIED MECHANICS. 467 

and tbey break with loads less than either /a or that given by 
(4). Curiously enough, however, when struts of the same 
section but of different lengths are tested, their breaking loads 
follow, with a rough approximation to accuracy, some rule as 
to length. Let us assume that as f =^f^ for short struts and 
what is given in (4 ) for long struts, then the formula 

^ Ml? 

may be taken to be true for struts of all lengths because it is 
true both for short and for long ones. For if I is great we 
may neglect 1 in the denominator and our (10) is really (4); 
again, when I is small, we may regard the denominator as only 
1, and so we have w =^f A.. We get in this way an empirical 
formula which is found to be fairly right for all struts. To put 
it in its usual form, let i = a z^^, k being the least radius of 
gyration of the section, then 



I 



(11), 



where a is 4//e tt^. 

If F does not act truly at the centre of each end, but at the 
distance h from it, our end condition is that y = h when x = I. 
This will be found to explain why struts not perfectly truly 
loaded break with a load less than what is given in (4). 
Students who wish to pursue the subject are referred to 
pages 464 and 513 of the Engineer for 1886, where initial 
want of straightness of struts is also taken account of. 

When we consider, therefore, how the rule (11) has 
been arrived at, it is evident that it needs to be tested 
by practical experiment, the constants a for various materials 
and k- for various kinds of section being also determined by 
experiment. This has been done, and on the whole we 
feel fairly well satisfied when the rule is put in the following 
form : — For a strut ivhose ends are hinged, or a column tvJiose 
ends are not fixed, the breaking load in pounds is equal to the 
breaking stress per square inch given in Table IX. multiplied 
by the area of cross section in square inches, and divided by 
1 + nB where n is given in Table XI., and b is given in 
Table X. 



468 



APPLIED MECHANICS. 



E:rercise for Advanced Students. — Test in the six cases of 
Table XL to what extent this rule agrees with (11). 



TABLE IX. 



Breaking Stress, in pounds 
per square inch. 



Cast Iron ... 
\Yrouglit Iron 
Timber 



80,000 

36,000 

7,200 



TABLE X. 

Value o/b for struts of the sections shoivn in Table XI. The first column 

gives the length of the strut divided hij its least lateral dimension. 



Lengtli divided "by 
Lateral Dimension d 



Bfor 
Cast Iron. 



B for 
Wrought Iron, 



B for Strong Dry 
Timber. 



10 


0-748 


0-132 


1-0 


15 


l-6.« 


0-300 


3-6 


20 


300 


0-532 


6-4 


25 


4-64 


0-832 


100 


30 


6-76 


1-200 


14-4 


35 


9-20 


1-632 


19-6 


40 


12-00 


2-132 


25-6 


45 


18-72 


3-332 


40-0 



EXEECISES. 

1. Find the breaking load for a cylindrical strut of wrought iron 
3 inches diameter and 10 feet long, supposing it to be not fixed at the 
ends. Ans., 30 '1 tons. 

2. A hollow cast-iron pillar 10 feet long, and fixed at the ends, has an 
external diameter of 6 inches ; what shoidd be the thickness of the metal 
to carry a load of 30 tons, allowing a factor of safety of 8 ? Ans., -42 in, 

3. What is the safe load for an angle iron, least breadth 3 inches and 
7^ feet long, acting as a strut, firmly fixed at both ends ? Factor of 
safety, 6. Ans., S'ol tons. 

4. The diagonal brace of a "Warren girder is 10 feet long, and is 
composed of two tee-bars 6" x 3" x J", placed back to back and 
riveted together. Find the maximum compressive working load which 
may be applied to it when the ends are firmly riveted to the boom. Use 
a factor of safety of 4. Ans., 11 b tons. 

5. Section 4, Table VI., has flanges 4'02 inches broad, -56 inch 

thick, depth over all 8 inches, web -^2 inch thick. The greatest moment 

of inertia about a line through its centre of gra\-ity is 74*2, according to 

the rule of the table. Its least moment of inertia is about a line at right 

, ^ ,, ^ , , . -56 X (4-02)3 e-ss X (-42)3 
angles to the first, and is -~ x 2 -t —^ -, or 6*1. 



APPLIED MECHANICS. 



469 



The area of the section is 7*39 square inch. The ultimate stress being 
taken as 45,000 lbs. per square inch, and the Young's modulus as 
30 X 10^ lbs. per square inch, the following are the breaking loads, 
according to Euler's theory, if it is the section of a strut fixed at the ends 
of the following lengths. For these loads to be withstood, it is necessary 
to carefully adjust the application of load at the ends so as to have 
absolutely no beRding until the breaking loads are reached, w = a/ or 
4 E I tt^/l^, if L is whole length of strut fixed at the ends ; the lesser 
answer of the two to be taken. 



L in inches 


96 


120 


144 


180 


216 


240 


288 


Brea,ldng load in tons 


148-5 


148-5 


148-5 


99-5 


69-1 


56 


38-9 



TABLE XI. 
Values of n for struts and pillars of the following sections :■ 



Square of side d, or rectangle with smallest side d ... \ 



Value 
of 71. 



^i 1-00 



Hollow rectangle, or square with thin sides 

Circle, diameter a 

Thin ring, external diameter a , 

Angle iron, smallest side <^ 

Cruciform, smallest breadth, f? 




2'00 



If we want the breaking load for a strut whose ends are not 
hinged, it is necessary to find in what way it tends to bend, and 
to use the above rule regarding the strut as hinged at two 
points of contrary flexure. Thus in Fig. 270 the strut or column 
B is as strong as a stmt hinged or rounded at both ends, whose 
length is only a h. The rule becomes — For a strut fixed at both 



470 APPLIED MECHANICS. 

ends^ calculate by the above rule, but take n one fourth of what 
I have given in Table XI. For a stmt one end of which is 
fixed and the other is only hinged ; calculate the breaking load 
as if both its ends were hinged ; then calculate as if both its 
ends were fixed, and take the mean value of the two answers. 

373. Struts with Lateral Loads. — If the lateral loads are such 
that by themselves, and the necessary lateral supj)orting forces, they 
produce a bending moment which we shall call ^ [x), then (1) Art. 

372 becomes F 2/ + <p [x) = - ei~^. ^^^len we know the lateral 

loads we know (p (x), and it is quite easy to integrate. Thus let a 
strut be uniformly loaded laterally, as by centrifugal force or its 
own weight, and then <p[x) — }^iv {I - xf' if iv is the lateral load 
per unit length. We find it shghtly more convenient to take 

<p {x) = \^ I cos. — X, where w is the total lateral load. This is 

not a very different law. Hence 

dhj r . 1 w^ IT - ... 



We find here that 

iw^ 

21 



cos. ^jX (2). 



xu 1. n J E I TT" \ , . , 



Observe that when f = this gives the shape of the beam. The 
deflection in the middle is 

Vi = ^> (3), 

and the gTeatest bending moment /x is ^u = f y^ + ^ w ?, or 

If w = 0, and if fi has any value whatever, the denominator of (4) 
must be 0. Putting it equal to 0, we have Eider's law for the 
strength of struts, which are so long that they bend before break- 
ing. If Euler's value of f be caUed u, or u = e 1 7r^/4 1^, (4) becomes 

If 2g is the greatest distance of a poiut in the section fixm the 
neutral line on the compressive side, or H i -r- z^ = z, the least 
strength modulus of the section, and a is the area of cross- section, 
and if / is the maximum compressive stress to which any part of 
the strut is subjected, ^ + - =/. Usmg this expression, if fi 
stands for - (that is, Euler's breaking load per square mch of 



APPLIED MRCHANICS. 471 

F 

section), and if w stands for - (the true breaking load per incli of 

A 

section), then , „. , „. ^; 

This formula is not difl&cult to remember. From it w may be 
found. 

EXERCISES. 

1. Every point in an iron or steel coupling rod of length 25 inches 
describes a radius of r inches. Its section is rectangular, d inches in the 
plane of the motion, and h at right angles to this. We may take 
w = Ibdrn^ 4- 62,940 in pounds where n — number of revolutions per 
minute. Take it as a strut hinged at both ends for both directions in 
which it may break. (1) For bending in the direction in which there is 

no centrifugal force where i is —x-. Euler's rule gives -ttt^o-- Now we 

shall take this as the endlong load, which will cause the strut to break in 
the other way of bending also, so as to have it equally ready to break 
both ways. (2) Bending in the direction in which bending is helped by 
centrifugal force. Our w is the above quantity di^ided by hd^ or, taking 

E = 3 X 107, w = 6-17 X ^ X 106. Taking the proof stress / for the 

steel used as 20,000 lbs. per square inch (remember to keep/ low, because 
of reversals of stress), and recollecting the fact that i in this other 

direction is -—^ we have (6) becoming 

8-4 X 108 (i _ 308 ^'\ ( 1 _ ^A = n^r^r -^ d (7). 

Thus, for example, i£ 5 = 1, ^ = 30, r = 12, the following depths d inches 
are right for the following speeds. It is well to assume d^ and calculate n 



from (7). ^ 



1-5 I 2 



2-0 



3 



4 



6 



n 205 1 277 327 368 437 545 

2. A roimd bar of steel, 1 inch in diameter, 8 feet long, or / = 48 
inches. Show that an endlong load only sufficient of itself to produce a 
stress of 1,910 lbs. per square inch, and a bending moment which by itself 
would only produce a stress of 816 lbs. per square inch, if both act together 
produce a stress of 23,190 lbs. per square inch. 

Students will find that this subject will well repay further 
study. The effect of a small lateral load on a strut is sometimes 
very striking. Again, in tie-bars it is very important to consider 
the effect of a lateral load. The subject is treated more fully in a 
paper in the Philosophical Magazine^ March, 1892. 

374. Beams without Compression. — In Art. 368 we have 
seen that a tensile load applied to extend a beam may not only 
diminish the greatest compressive stress, but also the tensile 
stress. Again, there are many cases of beams or infinitely 



472 



APPLIED MECHANICS. 



flat arches in which there is no tensile stress anywhere, lu 
such a case, of course, the earth takes the necessary tensile 
load. When the pneumatic wheel tyre was invented, Pro- 
fessor Fitzgerald pointed out to me that columns to support 
loads, and military bridges easy to pack -and unpack, might be 
made of inflated tubes, the solid material being everywhere in 
tension. I consider this a notion of great importance. In a 
thin straight tube of circular section if the greatest bending 
moment is M and R is the radius, t the small thickness of the 
material, the compressive stress anywhere due to bending is 

— y where y is the distance from the diameter which is the 

neutral line of the section on the compressive side. The 
greatest compressive stress is m/tt r'^ t. Now imagine the tube 
to be subjected to internal fluid pressure p above that of the 
atmosphere ; there is a tensile endlong stress P7rR~-f27rR^or 
p r/2 t, and hence the greatest compressive stress is m/tt r- t 
- P r/2 t. This is just o when p = 2 m/tt r3. The greatest 
tensile endlong stress is then, of course, p r/^ ; but this is equal 
to the lateral tensile stress which the mere internal pressure 
produces. When, therefore, the internal pressure is just 
sufficient to remove all compressive stress in the material, the 
tensile stress, where it is greatest, is the same in all directions 
and is 2 MyV r~ t. W^e see, therefore, that great loads may be 
carried by inflated tubes of thin material if they are only large 
enough in diameter, or by a bundle of small tubes. Shearing 
forces might be taken up by side frames\'orks like lazy-tongs. 
I make no attempt here at exact theory. What I give is 
sufficiently correct to show the general value of the suggestion. 
One may go far in speculation on this idea — rigidity gamed by 
using thin material and subjecting it to internal fluid pressure, 
so that there shall be no compressive stress. The great ships 
of the future may owe their stiffness and strength to the 
general use of fluid pressure in those parts of them where 
cargo is stored, and the same pressure which gives strength 
may serve to keep out the sea in case of a leak. It is the 
means by which the leaves of plants are made rigid. 
Similarly, large flat areas might be made of considerable size 
by fastening together two plane sheets by means of many 
connecting ties so that they may not balloon out, and then 
inflating them like an air cushion. Aeroplanes of sufficient 
size to support a man by Lilienthal's method can be made with 



APPLIED MECHANICS. 473 

comparatively small internal fluid pressures and are not liable 

to make splinters when tliey fall to the ground, these splinters 

being a cause of considerable risk with aeroplanes made with 

sticks as stiffeners. Kites much larger than those suggested 

for military purposes might be made, in which the whole kite 

might be like an air cushion, or thin tubes with compressed air 

might take the place of the present bamboo framework. The 

inflation might be maintained automatically. 

375. Inflated Columns. — Again, a thin tube of radius r and 

thickness t has to act as a column carrying a load w, and this 

is the load which is carried when there is no axial tensile 

stress. The pressure of the fluid inside being p, we have 

TT r2 p = Y>r . . . . (1). Also the lateral tensile stress produced in 

w 
the material is p r/^ or ■ -^ so that great loads may be sup- 

ported by inflated tubes of thin material if they are large 
enough in diameter. Thus, for example, I find that a tower 
of thin steel 1,000 feet high would have in it a lateral 
tensile stress of only 3 tons to the square incli, due to its 
own weight and the necessary fluid pressure. Being all in 
tension there is no danger of instability such as exists in 
ordinary pillars. If large in diameter, the hemispherical top 
cap becomes of importance as a load. Any moderate 
diameter like 20 feet would bear many tons on the top in 
addition to the weight of the structure itself. Thus, 1,000 
feet high and 20 feet in diameter and -01 foot thick would itself 
weigh about 125 tons. Its hemispherical cap would weigh 6 '3 
tons, and it would support 325 tons on its top. The internal 
pressure would be 23 lbs. per square inch and the tensile stress 
10 tons per square inch. There would be no compressive stress. 

Neglecting lateral stifihess, whether we assume the adiabatic 
or constant temperature law for increasing pressure down- 
wards in the fluid, we are led to rules as to the relationship of 
radius of column to thickness of metal if the column is to have 
the same stress in its material at all levels. In fact, Fitzgerald's 
idea gives rise to a number of easy and interesting problems 
which I have worked out, but for which I have no space. 

In all probability it would be found cheaper to use 
long stays from the top, and possibly from several places at 
different heights to resist lateral motion due to wind pressure, 
etc., than to stiffen as with lazy tongs the sides of the tube 
itself. It is certain that the thing is practical. 



474 APPLIED MECHANICS. 

376. Tapering Column of Circular Section. — If a; is the distance 
down-svards to any section where t is the thickness and r the 
radius, j; then being the internal pressure due to fluid and q the 
external pressure due to the atmosphere ; if /^ is the compressive 
stress in the material and ic'^ the weight per unit volume of the 
metal and w of the inside fluid, 

lx.2Ttrt . wi + 7r7-2 . So; . w + 2 irr.Sr.g' = 5a; . — {pirr"- + 2 irrtp). 

It follows from this, if w = cjj'^ly, as w^ is constant and ^ is a 
known function of cc, that 

d/- , 0' ( dp ., , X , 2 ?yi ) , ^ , X ri 



dx 2 ULv^ f 

if « =z 1 + '^f^if, where / is the greatest stress in the material ; 

P =2h (1 - h^y, ? = S'o (1 - ^^)^' 

where s = — — . taking the same fluid inside and outside. The 
7 - i 

result is, if <^ is taken to he small, 

2_!^^ X -I- ^^ + ^^ ^ log. (1 -bx) ■\- 2 log. T = constant. 
af a 

If ^ is not small, the integration seems troublesome. 

377. The limiting length of a vertical prism supporting its own 
■weight. — h being measured downwards to any place from the 
centre of gravity of the uppermost section, let i/ be the deflection. 

Let the load per unit length be w, and let w = I w . dh, the total 



= I w . dh, 



load above any section. Considering bending moment m and m + 5m 
at the sections at h and /i -\- 5A, we see that w\5y + J ic.5h.di/=^5 m, 

80 that w =-—=—-. —. Also M = - E I -^'. 
di/ dh dy dx- 

In the most general case, where m and i vary, and where there 
is a load f on the top, 

(J...* + .)|=-4;g-..g....o). 

Let I and m be constant, and let f be 0, 

f|+!l;, | = o....(2), 

dh^ EI dh 
an equation whose solution is useful in other problems. If h is 
the total length of the column whose end is fixed in the ground, 
lot X = hjn and (2) becomes 

^^=-»,4....(3), 
dx^ dx 

3 

EI 

The solution of (3) in series may be indicated by 
t/= axy(x) -\- B x\f (rr) . . . . (4). 
If we put -r^ = when a; = at top, we find s = 0, and the 



APPLIED MECHANICS. 475 

other part of (4) is 

^-^''' I ^"273-4 + 2. 3. 5. 6. 7~2. 3. 5. 6. 8. 9. 10 +'^''-} 

In this, if we put -- = when x = 1, we find 
^ dx 

= 1- + — + etc., 

2.32.3.5.6 2.3.5.6.8.9^ ' 

and by trial we find that this is satisfied hy m = 7 '85.* Hence 
if bending- occurs, wh^/ei = 7'85; so that this gives us the 
limiting height h of a uniform column. Thus we find h in 
feet = 8 X lOe/L^ for thin steel tubes, and h = 4 x IO^/l^ for 
solid rods of steel, where l is the ratio of length to diameter. 
Thus, for example, the maximum height of a tube 8 feet in 
diameter is 800 feet. 

Let A ^2 := J . ^ = diameter in inches. Greenhill gives 

p is ^. First put j^ = x^z, then put x^ = r'^ ; we get 
dr'^ dr 



dx' 



''''-^■t^&''-'> = '---<^'^- 



This is 



:j5 + r^^ + (F.^-.a. = 0....(3), 
dr^ dr 



4 '}V 
Bessel's equation, where F = ^ ,^ , n- = 1. Solution of (3) is 

2 = AJ^ (Arr) + bj_^(At). 

Consequently 

i? == x/x [ A J J ikx t) + B J- 3 (kx t) I . 

_ = when x = 0, makes a = ; 

.'. p = B V'x J _ 1 (A-.1; t) . . . . (4). 

Put « = when x = k, the lowest point. J _ 1 (^'/^ '^) = 0. H 

c is the least i-oot of j_jl (c) = 0, then c = kh^. Greenhill gives 
the expansion of j _ 1 (c), and finds root by trial. 

378. Stability of Shafts. — Rankine considered the effect of cen- 
trifugal force on shafting in 1869. Professor Greenhill has investi- 
gated the stability of a shaft between bearings (see his Paper to the 
Institution of Mechanical Engineers, 1883). He took into account 

* Prof. Maurice Fitzgerald piiblished curves showing the values of y and 

-^ when X = 1 for various values of m (Proc. Phys. See. of London, Oct., 

1892). Prof. Greenhill had previously given the solution, not only for 
uniform solid cylinders, but also for cones and paraboloids of revolution 
(Proc. Camb. PhU. Soc, 1881). 



i76 APPLIED MECHANICS. 

an end thrust, f, a twisting moment, x, and C(?ntrifugal force. He 
found in practical examples of propeller shafts that the twisting 
moment was not nearly so important as the thrust in determining 
the maximum length of shaft for stability ; in fact, that the shaft 
might he designed, merely like a strut, by Euler's formula. I 
consider, therefore, that such a shaft ought to be studied under 
the rules of Art. 372. 

379. Whirling Loaded Shaft. — If a imiform shaft, originally 
sti-aight, is w lb. per unit length, and has an angular velocity of 
a radians per second ; if b is its flexural rigidity, or e i ; if the 
deflection from straightness is y at a point distant x from the 

centre, then the load due to centrifugal force is — ah/ per unit 

length. If we also take into account the effect of gravity, there is 
one position in a revolution when the centrifugal force and the 
weight produce their greatest effects. We may approximately 
take it that the path of every point is a circle, and that the weight 
is a radial force always acting in the same direction as the centri- 
fugal force. If we take into account also an 'endlong thrust, f, we 
are led to the equation 

d^y ^ F d^// ^' «^ ,, «■ 



d.> + l,17^-lFJ^-^ = '>----('^- 



and to the solution 
y = Ai sin. /Sir + A, cos. j8a: + A3 e y'' + A4 e~y ^ — glo? ... - (2) 



I'here 



VF^ icaA 



'^ 2b "^ ^\/ 4 b- ' gji 



(3). 



y is to have the same values for equal positive and negative values 
of X. so that A] = 0, Ag = A4. Putting y r= where x=zl and con- 
fining our attention to shafts whose bearings do not constrain 
them in direction at the ends, so that drijjdx'^^ where a: = / we 
find Ai = 0, A, = -fglcC" (72 + 0^ COS. )8>, A3 = a^ = 0'gi2a^ (7^ + 0^ 
cosh yl. 

The bending moment anywhere is e i dr^yjdx'^, and hence the 
bending moment in the middle, where it is evidently greatest, is 
Mo = E I ( - )8- A2 + 2 A3 72), or 

=> 3V. (3^, - 3^0 /"^ (^ + ^°-) • • • ■ «• 

The oTeatest stress at the middle is 

where z is the strength modulus of the section, and a is the area of 
the section ; or if r is outside radius of a circular shaft we have 

^= ^"'^ h^l - ^l) /v/^-^ + 4^«^«/^ + F/a . . . . (6). 



APPLIED MECHANICS. 477 

Exercise. — Show that in the case of a propeller shaft 13-4 inches 
diameter, of length 98 feet between its hearings, with endlong thrust 
50,000 lbs., if a = 2 TT, or 60 revolutions per minute, the important terms 

in the above calculation are p2^4b2=30 x 10 "■'■*, and wa^Jg-&=^ 

89 X 10 ~ . So that the centrifugal force is 300 times as important as 
the endlong thrust. 

If we altogether neglect twisting moment and endlong thrust 
and write n^ for w o?jg b, we have to solve 





d^ XI . IV 
dx'^ -^ B 


0... 


.(7). 




Our old /3 = 7 = 


«, and the greatest 


stress 


is 




/• 


/I 1 


v^^'' 


B «2 . . . , 


.(8). 


/= E WK 

\cosh nl cos. 


This is infinite if 


cos. nl is 0, that is : 


If nl = 


= f;that 


is, if 




-^^Cwf 





(9). 





This limiting length may be obtained very simply by leaving ouc 
the constant term in (1), as Rankine does. 

Mr. Dimkerley (Phil. Trans, for 1894) discusses the stability 
under centrifugal force of shafts loaded with pulleys, and he has 
illustrated his results experimentally. 

EXERCISES. 

1. If the critical length of a shaft 13-4 inches in diameter, subjected 
to endlong thrust alone, is equal to tiie critical length when subjected to 
centrifugal force alone, show that r = 68,500 a. Show that if the length 
is 98 feet, the critical f is 327,600, and the critical a is nearly 4'1 radians 
per second, or 46 revolutions per minute. 

2. Professor Greenhill's result is that if f is the endlong force, and t 
the twisting moment, the limiting length of shaft being 2 /, we have 



^2 



p T- 

+ 



4Z2 EI "^ 4e2i2" 

The propeller shaft of the Cimard steamer Servia is of wrought iron, 
22^ inches diameter. The pitch of screw is 35^ feet. At 53 revolutions 
per minute the indicated horse-power was 10,350. Assuming that all 
this power is really iitilised — an assumption which is, of course, quite 
wrong— prove that r = 181,530 lbs., and that t = 12*3 x 10^ pound- 
inches. Take e = 29 x 10", and show that the above formula becomes 

^ = 4-98 X 10-' + 2-8 X 10-^°, 

so that the twisting moment term is quite inconsiderable compared with 
the thrust term. Show that 2 ^ = 4,454 inches. Consider this shaft of 
22J inches diameter, 4,454 inches long between bearings, subjected to no 
thrust or twisting moment, and not even to its own weight ; prove that it 
cannot be rotated at a higher speed than a = "14, or \\ revolution per 
minute, without fracture by centrifugal force. 



478 



CHAPTER XXII. 

METAL ARCHES. 

380. The student must examine drawings and actual specimens 
of "bridges to see how tlie weigM of a roadway is brought to bear 
upon the arched ribs whose fimction is to carry weight, trans- 
mitting it wdth certain horizontal forces to the abutments. Our 
problem is, given the distribution of vertical loads on an arch-ring 
of which the shape of the centre line and the shape of cross-section 
everywhere are known, to find the stress everywhere. If k z g is 
the centre line of the rib shown in Fig. 270, then any cross-section 
E c is supposed by our theory to remaia plane. The resultant of 




Fig. 270. 



aU the loads acting to the right of b c, together with the resultant 
force R at the abutment, is a force whoso direction is shown at d 
by the direction of the line of resistance there, and the force 
polygon shows its amount. We saw that in Fig. 261 this force, 
which we there called r, produces a bending moment ii at the 
section whose amount is p . o d. But evidently this is the same as 
F . oj (Fig. 271) if o J is perpendicular to f. A much more 
imjDortant thing for our present purpose is to know that it is also 
equal to the horizontal component h of the force f multiplied by 
the vertical distance o l. This the student will easily prove for 
himself. 

Now if all the loads are vertical, we know from Art. 349 that 
(1), the resultant force f at every joint has the same horizontal 
component h as at any other. This is represented on the force 
polygon (Fig. 236) by oh. It is, of course, also the horizontal 



APPLIED MECHANICS. 



479 




thrust on each of the abutments, and is generally called the 
horizoDtal thrust of the arch. Of course the force polygon shows 
all the forces r at all the sections. (2) If two vertical lines be drawn 
through K and g, as in Fig. 270, and if any line of resistance be drawn 
whose ends k' and g' are in these lines, and 
if k' and g' are joined, the vertical height 
anywhere, i> T,of this link polygon is inversely 
proportional to the o h of the force polygon 
by means of which it is drawn ; so that if 
two such figures as k'z'g' are drawn, the 
vertical ordinates dt are all in the same 
proportion. We see, therefore, that since 
the banding moment at b c is the horizontal 
thrust mrdtiplied by o l, the vertical distance 
from the centre line anywhere to the line 
of resistance there represents to some scale 
the bending moment at the section there. 
At V and u the lines cross. These are points 
of no bonding. From k to u, and from g to v the bending 
moment tends to make the arch more convex upwards. From 
V to u the bending moment tends to make the arch less convex 
upwards. If we can 
only find the true line 
of resistance li' z' g ', 
we know by Art. 129 
the stress in every 
section. The true po- 
sition of k' z' g' is 
determined by these 
conditions : — 

1. In an arch 

hinged at the ends, if k and g are the centres of the hinges, 
we know that the line of resistance must pass through them. 
We only need one other condition, and that is given by the 
statement : the yielding everywhere of the arch must be such that 
the distance k g remains constant. 

2. In an arch fixed at the ends, if the fixing is perfect, we have 
the above condition that the distance k g remains constant, and 
also the condition that the inclinations of the centre line at k and 
G remain constant. Just as in the case of beams fixed at the ends, 
it is exceedingly difficult to fix the arch at the ends so perfectly 
that we can rely upon this condition being fulfilled ; and hence, on 
accoimt of our uncertainty, we prefer to use arches hinged at the 
ends. 

3. The simplest case of all is that in which there are three 
hinges in the arch, one at each abutment k. and g, and another at 
the crown. Even when the loads are not vertical it is easy to find 
the line of resistance, because two corners and a point in it are 
given, and we have merely the exercise of Art. 106. 

Any system of oblique loads is given, acting upon an arch 
whose end hinges are k and g and whose crown hinge is l. 
Find the resultant of all the loads from k to l, and let it be a b 




480 



APPLIED MECHANICS. 



(Fig. 273). Find the rcsuliaut of all the loads from l to a, and let 
it be BC. Draw ab and b c in the force polygon (Fig. 274). 
Choose any pole o. Join o a, o b, and o c. Draw k p, y q, and o, r 
parallel to o a, o b, and o c. The resultant of a b and b c passes 
through s. Now join o s and draw the new diagram, c o' parallel 




to G 8, and join o' b ; so that ic p, r of, o,' g correspond to o' a, o' b, 
and o'c. We have then a link polygon. But it does not pass 
through L. Produce p q,' to meet k g in t. Join t l and let it meet 
the gi^^en forces in p" and q". Join k p" and o," g, and k p" q" g is 
the link polygon or line of resistance required. I set the problem 
as an exercise for students, and one of them, Mr. Stansfield, gave 
me this solution. 

If the loads are vertical, and k, g, and z are the given hinges, Fig. 
272, draw any line of resistance k z' g', its corners z' and g' being in 
the verticals from z and g. k z" g' is a straight line. Draw a vertical 
2.' z X z" through z. C onstruct a figure on the base k g such that 
if b' is any point in it, b' c is in the same proportion to b d that 
zx is to z' z". This will evidently be the true line of resistance, 




Fig, 275 



and the vertical ordinates between it and the centre lines of the 
portions of the arch will be the bending moments. One example 
of each of these must be worked out by each student. 

381. Arch Hinged at the Ends.— Imagine the small slice of 



APPLIED MECHANICS. 



481 



the arch between two sections b o c and b' o' g', at the 
small distance 5s asunder, to yield, and study the effect of 
the yielding of this portion alone, exactly as we did with 
our spring in Art. 370. That is, we imagine the part o' g 
(Fig. 275) to he absolutely rigid and fixed ; the part o k to be 
rigid, but to move about o as centre, as if ok were merely 
a pointer, the motion k k." being the angular motion of b c 
relatively to b' g', multiplied by the straight distance o k. The 

angular change from o to o', if o o' is 5s, is — 5s, if m is the 

bending moment, i the moment of inertia of b c about o through 




its centue of gravity, and e is Young's modulus; so that kk" 

is — 5s . o K. The horizontal component of this is k k'" or 

K k" cos. K." K k"'. But this angle is the same as k o n, whose 
cosine is o n/o k. Hence the horizontal motion of k due to the 

• ij- 4!+i, ^'4.^^ T . M.Ss.GK ON M . O N . 5s ,,, 

yieldmg oi the little slice is • — or .... (1). 

£ I OK. !E I 

A.nd as k does not yield at all, we make this sum zero. We beg 
to point out that we cannot in the same way state the vertical 
displacement of k. If o n is called i/, and if k n is called ^, then 

(1) is 2 — 3/ . 5« = ; and we might in the same way think that, 

as — a; . 5s is the horizontal displacement of k due to the yielding 

of the slice ; therefore the sum of all these terms ought also to be 0. 
If the end G is fixed, this is tru.e, and it enables us to calculate the 

fixing moment. But in truth 2 — a; . 5s is not if g is hinged ; 

it is equal to the angular movement at g multiplied by k g. Note 
that for the horizontal displacements the angular yielding at o 
produces no effect. 



Q. 



4:82 APPLIED MECHANICS. 

Since m is represented to scale by the distances o l, (1) becomes 
2 '- '■ — = . . . . (2). In Fig. 'ilQ k z g is the centre line 

of the arch of which k and g are the hinges, k z' g is the line of 
resistance which we want to find. If a student wishes to keep (2) 
in his memory, let him imagine that vhe line K. z g has positive and 
negative density, represented by m/ei per unit length. Then (1) 
or (2) tells us that if the whole positive jDart of the linear mass is 
Wp, and if i/p is the distance of its centre of gravity from k. g, and 
if the whole negative part of the linear mass is m^, and if t/n is the 
distance of its centre of gravity from k g, then m.p y^ = m^ Vn. 
To find K z' G, let k z" g be any line of resistance through k g 
drawn at random. Then l n = Z; . p n, where k is an unknown 
constant ; and if we know k, knowing p j^, we can find l n. 
Divide the given centre line e: o z g into any number of equal 
parts, and draw an ordinate n o p at the middle of each of them. 
All values of 5s are now 

,,^, . _ON.OL - ,„, ^on(lx-on) 
equal, and (2) is 2 = . . . . (3), or 2 ^^ ' = ; 

80 that, as l n is A; , p n, we have k 2 = 2 .... (4), 

J 7 • -J XT ^0^2 ^ON.PN ... mi.- • M 

and k is evidently 2 -J- 2 .... (5) This is easily 

calculated. Thus, for example, take k. z q, an arc of a circle, span 
200 feet, rise 30 feet. For a given system of unsymmetrical loads, 
K p z" G was drawn. Dividing k z g into sixteen equal parts and 
raising perpendiculars at the middles of the parts, jve found the 
following values. The curves as drawn were measured in inches, 
no attention being paid to scale. The values of i are in inches to 
the fourth power. We see that k = -03014 ^ -07447 or 0-405. 
Multiplying each value of p n by this, we find the values of l n. 
The horizontal thrust corresponding to this true line of resistance 
is greater than the one for k z" g in the ratio 1 : '405, and its 
force polygon may be diawn. The student needs no further hints 
for the completion of the calculation. The loads taken were as 
foUows : — 



APPLIED MECHANICS. 



483 







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464 APPLIED iJtlCH.O^ICS. 

382. Arch Ring Fixed at the Ends. — In the case of a symme- 
tric 1 1 v'r:.. :J:e ^ni ~x'.z.2 r. ments l)emg called wij, we simply 
hav- hi; i^r^ I- :r ;u; : . :-a everywhere to the bending moment 
considered in the last case ; or, rath-rr, the hending moment at a 
place o is o L - mi, instead of bcing^ merely o Jm. (2) becomes 

S ""--"'-^-^ ^' =0....(2). 



^ OK.l.x-o>- ^, ^ ^ _ j3^_ 



(3) becomes 

and (4) becomes 

i2iil-i^"-2ii^-,„,2"-=0....(4). 

I I ^ I ' 

Bat we have another condition to satisfy. The angular change in 
the length Ss is S«, and the integral of this along the centre line 

fromxtoTisO. H- ce 2 ^^^-— ^ = 0, or 

*2^ - 2^ - »ii5^ = (5). 

When we find the summations in (4) and (5), it is easy to calculate 
the two unknowns n^ and k. When f ouni m^ will hie to the same 



Hg.277. 

scale as that to which the distances o l represent bending momeiit, 
and the curre of Fig. 277. x z' t. obtained when we know k must be 
displaced downwards everywhere by the constant displacement imj. 
till its appearance is that shown in x'm t'. A thoughtful student 

win see that the 2 — k x . 5* summation is also .... (6) in this 

EI \ / 

case, but that (5) and (2) cannot both be true unle^ (6) is also tme, 
because the loading is symmetricaL 

In the case of an nnsymmetrical loading, the bending moment 
efverywhere is l^s than o l by an amount which is m^ at the end x^ 
and is m^ at the end t, and which changes. A little consideration 
will show that we mu^^t lower the curve k z' g of Fig. 276 by 
amounts shown for each vertical line by the ordinates of snch » 



APPLIED MECHANICS. 485 

diagram as kk' g' g of Fig. 271. If a; is the horizontal distance of 
the of any section fi-om k, and I is the wliole span, then we must 

subtract the amount m^ + ^-^—j — ^ from the ol of Fig. 276. 

Call this ;»i + p ar, then (4) becomes 

o N . p N ^ o n2 on ^ o X . a: 

A-2 ^ 2~ - ;«,2— - P2-^— = 0.... (4). 

(5) becomes 

PN ox 1 ^ r. ,-s 

I I ^ I I ^ ' 

"We also now find another condition by stating that the sum of the 
vertical displacements of k is when g is held fixed — that is, 

2 — KX . 5s = . . . . (6), and this gives us 

,^PX.KX ^OX.KX ^x X- 

ic 2 2 "'i 2 p2— = 0....(6); 

I I ^ I I ^ " 

and the r-ummations indicated in (4), (5), and (6) being effected, we 
have no difficulty in solving for k, m^, and p, and so finding the 
true line rf resistance. 

The table thows these summations. It is for the same arch 
ring, with same loads, as in Art. 381, and therefore some of the 
columns are the same. 

The student will note that if t/^ is the ordinate of the centre 
line of the unloaded arch at any place, and if y is the ordinate of 
the leaded arch, and if ^^ - y, the downward motion of the point, 
be called z, then, imless in some case there is very great slope 

ui ieed, we may take t4 -^ ) 1 + ( /t ) ( " ^^ ^^® change of 

cm-vature. The proof of this is easy. Equating this to m'ei, we 
see that we have only to imagine i at every place divided by 

1 + ( -i- I [ - there, calling it by the new name i^, and we take 
— s = m/e i^. Having a diam-am showing evervwhere th-e value of 

m/e I, it is easy to obtain the diagi'am of z by graphical statics, just 
as in Fig. 217 we obtained the shape of a beam fi'om a diagram 

of M EI. 



486 



CHAPTER XXIII. 

MEASUREMENT OF A BLOW. 

383. In Aii:. 46 we considered what occurs when a chisel is 
cutting metal. We shall now consider the same subject from a 
point of view which at first seems difierent. Consider how it 
is that a blow of a hand hammer will indent a steel surface, 
whilst a steady force applied to the same hammer-head would 
require to be very great to produce any indentation. The 
pressure between the hammer and the steel is very great, 
and it must be all the greater because the time of contact 
is very short. Indeed, if a hammer weighs 2 lbs., so that 
its mass is 2 -=- 32"2 or -0621, and if, just before touching 
the steel, its velocity was 10 feet per second, then we 
know that its momentum was lOx'0621, or -621. Now, if 
one-ten-thousandth of a second elapses from the time of 
actual contact until the hammer's motion there is destroyed 
— that is, until the elasticity of the steel is just about to send 
the hammer back again a little — the momentum -621 is destroyed 
in -0001 second ; hence the average force (notice that it is a 
time average) acting between hammer and steel during this 
short time must have been '621 -^ -0001, or 6,210 lbs. It is 
certain, however, that this average force is less than what the 
force actually was for some very small portion of the time. 
We observe, then, that we cannot tell the average force of 
an impact unless we know two things — the momentum and 
the time in which it was destroyed. Now the duration of 
an impact depends greatly upon the nature of the objects 
which strike one another, and we see that the average force 
of a blow is less as the time is greater. Sometimes, instead of 
a great force acting for a very short time, what we require is 
a smaller force acting for a longer time. For instance, when 
cutting wood we obtain this result by using a wooden mallet 
and a chisel with a long wooden handle, because the force 
required to make the chisel enter the wood is not very great, 
and we wish this force to act for some time, so that much 
wood may be cut at one blow. In chipping, we have the time 
short, because considerable force is required to cause the chisel 
to enter metal. The duration of an impact depends on the 



APPLIED MECHANICS. 487 

shapes of the bodies and their masses, and on the elasticity of 
their materials. 

384. Why is it that ia driving a nail into wood our blows 
seem to be of no effi?ct unless the wood is thick and rigid, or un- 
less it is backed up by a piece of metal or stone "? It is because 
the wood yields quite readily, and so prevents the hammer 
losing its momentum rapidly. There are few subjects in 
which people are so apt to have erroneous ideas as in this one 
of impact. Thus a man will speak of the force produced by :i 
weight falling through a height without having any idea ol' 
the tiTYie during which the motion of that weight is being- 
stopped — in fact, without considering what time the weight 
is allowed for delivering up its momentum. Now, a little 
consideration will show that tlie mean force of the blow will 
be quite different according as the weight falls on a long and 
yielding bar or on a short and more rigid one. If we could 
imagine bodies to be formed of perfectly unyielding materials, 
then the slightest jar of one against the other would produce 
an infinitely great pressure between them ; and in the blow 
produced by a falling body there may be every gradation from 
exceedingly great pressures to very small ones, depending on 
the yielding power of the body that is struck. Everybody is 
acquainted with the sensation produced by suddenly placing 
one's foot on a level floor when one was preparing for a step 
downwards. The downward momentum of the body is 
suddenly destroyed, and there are great pressures in all the 
bones of the body. Carriages are hung on springs for the 
purpose of preventing their losing or gaining momentum with 
too great rapidity when the carriage wheels pass over obstacles. 
When we are sitting on a hard seat in a third-class railway 
compartment, and the carriage gets a slight jerk upwards, 
momentum is given much too rapidly to our bodies for perfect 
comfort, and to sit on cushioned seats is preferable. A cannon- 
ball is safely, because comparatively slowly, stopped by sand- 
bags or bales of cotton. 

385. Example. — A pile driver of 300 lbs. falls through a 
height of 20 feet, and is stopped during O'l second. What 
average force does it exert upon the pile % A body which has 
fallen freely through a height of 20 feet has acquired a velocity 
equal to the square root of 64*4 x 20, or 35-89 feet per second. 
Its momentum is 35-89 x 300 -^ 32-2, or 334-4, and this 
divided by 0-1 gives 3,344 lbs., which, together with the 



488 APPLIED MECHANICS. 

weight 300 lbs., is 3,644 lbs., the answer. From the iiLstant 
when the motion of the driver ceases to diminish, the force 
exerted by it is its own weight. We have considered the 
time average of the force. The average force of friction in 
pounds between the pile and the ground, multiplied by the 
distance in feet through which the pile descends during the 
stroke, is equal to 300 x 20, or 6,000 foot-pounds, if we neglect 
the loss due to vibrations of the body and the energy carried 
off to the ground to be wasted in earth vibration, and if we 
also igDore the fact that the weight really descends a little 
farther than 20 feet. The neglected energy may be the 
whole of the energy if, for example, the blow does not make 
the pile move further into the ground. We can only be 
certain about a time average force, and even in this case we 
must assume that there is an instant at which one of the 
bodies has the same average velocity in all its parts. 

385a. Example. — A column of water in a pipe 6 inches 
diameter, 30 feet long, was moving behind a piston at 15 feet 
per second. The piston's motion is stopped in 0-1 second. 
What is the time average of the pressure due to the stoppage 1 

Ans. The area of the circular section of the column of 
water being 28-274 square inches, the quantity of water in 
motion is 30 x 28-274 ^ 144 or 589 cubic feet, or 5-89 x 62-3 
or 366-9 lbs. ; and on the assumption that all the water has 
exactly the same motion, its momentum is its mass 366-9-^32-2 
multiplied by 15, or 1-70 units. The time average of the 
force required to stop the motion in O'l second is therefore 
(62-3 X 30 X 28-274 x 15) -f (144 x 32-2 x 0-1) lbs., which 
is equivalent to a pressure of 60-5 lbs. per square inch over 
the area of the piston. 

386. Suppose a body a to strike another b, and that we 
can neglect the actions of outside bodies upon them both. If 
A loses momentum, b must gain the same amount because 
their mutual pressures are equal and opposite during the time 
of impact. It is our knowledge of this fact that enables us to 
calculate the motions of bodies after they strike one another. 
Again, for the same reason, if from any internal cause the 
parts of a body separate from one another, either violently or 
gently, the total momentum remains as it was ; it is only the 
relative momentum which alters. Hence, when a shell bursts 
in the air, some parts move in the same direction more rapidly 
than before, but others less rapidly ; one part may double its 



APPLIED MECHANICS. 489 

velocity and another may drop nearly vertically, its forward 
motion being stopped, but, on the whole, the total forward 
momentum is what it was originally.* 

387. Examples. — If a cannon were perfectly free to move 
backward when the shot leaves it, the backward momentum of 
the cannon would be exactly equal to the forward momentum 
of the shot. Thus, if a shot of 20 lbs. leaves a cannon whose 
weight is 2,240 lbs. with a velocity of 1,000 feet per second, 
the velocity of the cannon backward would be 1,000 x 20 
-T- 2,240, or about 9 feet per second, neglecting the fact that 
the gases leave the gun also with a certain momentum. When 
a ship fires her broadside, each gun runs back, communicating, 
as it is stopped, its momentum to the ship, which heels over 
in consequence. A gun firing the above shot of 20 lbs. 
directly astern from a ship whose total weight is 600 tons 
gives to the ship (neglecting the momentum of water moving 
with the ship) so much momentum that its speed is increased 
(neglecting her friction with the water) 9 -^ 600, or "015 foot 
per second. We see, then, that a ship might propel herself 
by means of her guns. The steamship Waterwitch had power- 
ful steam pumps, wherewith she brought a great quantity of 
water in nearly vertically, and sent it out backwards on the 
two sides below water level. The momentum given to the 
water backwards was equal to the momentum given in the 
other direction to the ship. It is on this principle that Hero's 
steam-engine and Barker's mill work, the momentum given to 
jets of fluid passing out of certain pipes being equal to the 
momentum given in an opposite direction to the vessel from 
which the fluid passed. In all such cases the propelling force 
in pounds is numerically equal to the momentum of the fluid 
which passes out in a second. Thus, if from a vessel moving 
w4th a velocity of 14 feet per second water comes through 
orifices of 4 square feet in area with a velocity (relative to the 
orifices) of 20 feet per second, then the quantity passing out 
in one second is 4 x 20, or 80 cubic feet — that is, 80 x 62*3, or 
4,984 lbs. Now, recollecting that this water was first brought 
in and is now sent out, what is the velocity which we have 
really impressed upon it in the process ] At the beginning it 

* Of course the kinetic energies of the parts of the shell added together 
are greater than they were before the shell burst ; we are now merely 
speaking of the momentum. The total momentum of two equal bodies going 
in opposite directions with the same velocity is nothing, whereas their total 
kinetic energy is double that of one of them. 



4^0 APPLIED MECHANICS. 

was motionless with respect to the sea ; it now has a velocity 
of 20 — 14, or 6 feet per second with respect to the sea, so that 
the momentum given to it is its mass, 4,984^32-2 multiplied 
by 6, or 928*7 : hence, as this momentum is given every 
second, 928 '7 lbs. is the propelling force exerted on the ship. 
In one second the ship moves through 14 feet, so that the 
useful mechanical work done is 14 x 928*7, or 13,002 foot- 
])ounds. We have given to 4,984 lbs. of water a velocity of 
6 feet per second, the kinetic energy of this water is wasted, 
and this kinetic energy is J of 4,984-^32-2x6x6, or 2,786 
foot-pounds. In fact, we have altogether spent 15,788 foot- 
pounds, and 13,002 of this have been usefully employed, so 
that the efficiency of the method is 13,002 -^ 15,788, or -824, 
or 82-4 per cent. As a matter of fact, however, the friction 
in pumps and pipes usually causes a third of the actual horse- 
power given out by the engine to be wasted, so that the 
true efficiency of this method of propulsion is two-thirds of 
the above, or 0-55, or 55 per cent., neglecting the friction 
of the engine itself. You will remember a fact which has 
come in casually here : if the water leaves any turbine, 
water-wheel, or any propeller of a vessel with a velocity 
relative to the still water into which it passes, or if it has 
any other form of energy, this energy has been wasted. 

388. By calculation you will find that, when two free and 
inelastic bodies strike, the momentum communicated from one 
to the other is their relative velocity multiplied by the product 
of their masses and divided by the sum of their masses, and 
this quotient divided by the time of the impact gives the mean 
pressure. This pressure acts equally on both, of course, but it 
may not hurt both equally. If the bodies are surrounded by 
water, like ships, they can no longer be regarded as free 
bodies, and it is not easy to say in a few words how much 
mass we must add to the bodies to represent the mass of the 
water, which has also to undergo change of motion. In the 
case of a ship, the mass of water to be moved broadside on is 
much greater than when the ship is struck stem on. 

389. A body falling into a liquid sets it in motion, and 
this motion appears at distant places more and more nearly 
instantaneously as the liquid becomes more and more incom- 
pressible. The nature of this motion is known to us if we 
know the velocity of yielding at the place of contact, and 
from this the total momentum given to the liquid. Thi?' 



APPLIED MECHANICS. 491 

represents a very considerable pressure applied at the place 
of contact, and this pressure becomes greater as the velocity of 
the body, before it touches the liquid, increases. Hence a 
cannon-ball fired at sea rebounds from the water as from a 
rigid body. Hence also a man diving unskilfully, as he falls 
prone on the water, gets a very unpleasant shock, whereas a 
skilful diver enters in such a way as to make the momentum 
of the moving water as small as possible, and to make the 
creation of this momentum gradual. 

390. If a body of inertia or mass M with velocity v overtakes 
a body of mass M^ with velocity v^, the motion being in the 
same direction, there is an instant when they move at the 
same speed, and then they usually separate. The equal and 
opposite forces equalise their velocities until they are both 
moving with the velocity v, such that 

(m + M^) V = M v + Ml yi . . . . (1), 
or 

V = 'Il±J^Yl .... (2). 

M + Ml ^ ' 

There has been a communication of the momentum, M {y—v), 
or m1 (v — yi), and this is the amount of the impact or the 
time integral of the force from either on the other. We 
usually imagine the contact surfaces to be normal to the 
direction of motion.* During the impact the total kinetic 
energy, 

Ei = 1mv2 + jMiyi^ (3) 

becomes lessened to 

E = J (m + Ml) ^2 (4)^ 

the amount b^ — e being stored as strain energy. In truth, 
much of it travels off and vibrations take place (see Art. 486), 
but let us speak of E^^ — E as the stored energy. Assume 
that the energy of the bodies when free after collision 

* If not normal we may speculate on the connection between the ideas of 
force as rate of communication of m.omentum, the direction of the force 
being the same as that of the momentum communicated, and yet the 
direction of communication being different from both. Students who have 
leisure will find three quite neglected letters in Nature for 1878, by Prof. 
R. H. Smith, which will give bem novel ideas on the subiect — ideas likely 
to be of use to engineers. 



492 



APPLIED MECHANICS. 



1 



3 . _ -, -.2. . 



(m u + M^ u^ ) is less than e^ by an amount which 
is a fraction of the stored energy, or 

El — E2 = ^ (Ei — e) . . - . (5), 

where ^ is a constant depending on the nature of the materials 
and the shapes of the bodies. We know also that 

V = ; — -. — .... (6). 

From (2), (5), and (6) we can calculate 13, u^ and v in terms 
of the initial velocities, and we find that 

ui - u = (V - vi) yT^k. 

Or, " the relative velocity of separation is ^ 1 — k times the 
previous relative velocity of approach." This was Newton's 
assumption. It is usual to denote the ratio ^ 1 — ^ by the 
letter e, and to call e the " elasticity," but this is unscientific. 

391. Newton found by experiment that e = f.for com- 
pressed wool, iron nearly the same, ^f for glass. Hence for 
these substances our k has the values -^ and J. That is, in 
the compressed wool or iron -^ of the stored energy is wasted ; 
in glass only -J- of the stored energy is wasted. It is very 
difficult for us to imagine how this difference between iron 
and glass should occur ; and there are no recent experiments. 
Note that A; = 1 or e = indicates the case of maximum loss 
of energy ; in fact, there is maximum loss of energy when the 
two bodies continue to move together as the bullet and bob of 
a ballistic pendulum do. In this case the kinetic energy 
before collision is ^ m v^, and after collision it is 

i (M + Ml) ( "^^+y )'' or 1 (M V + Ml Vl)V(M + Ml). 

Suppose v^ = 0, or we have m impinging on m^ at rest, the 
energy remaining is f m v^ / ( 1 + — ). Hence, the greater M^ 

is in comparison with m the greater the loss. In fact, 

lost energy Mj 

remaining energy m 
So that the larger the stationary body before collision, the 
greater the loss. 

392. Example. — Thirty gallons of water per second enter 
a wheel in a direction A b from a horizontal pipe 4 inches in 



APPLIED MECHANICS. 493 

diameter, the shortest distance from the axis of pipe to vertical 
axis of wheel being 1*3 feet. The water leaves the wheel in a 
horizontal direction, c d, with a velocity of 3 feet per second, 
the shortest distance between c d and the axis of wheel being 
0*8 feet. What is the turning moment? And what is the 
power if the wheel makes 150 revolutions per minute? 

Ans. The 30 gallons, or 300 lbs., or ^^ cubic feet, have 

...,,. 300 144 
an initial velocity of ^^75 • 73 -"fi^A. ^®®^ P®^ second, and 

300 
mass ooTn- The product is the momentum [)er second, and is 

force. Multiply by 1-3, and we have the moment of the force 
due to the entering water, or 6 68 -2 pound-feet. The same 
mass multiplied by 3 and by 0*8 gives 22-36 pound feet; and 
subtracting this from the former, we have the resultant 
moment (or moment of momentum per second), 645*8 pound- 
feet. Multiplying this by 150 x 2 tt radians per minute, and 
dividing by 33,000, we find 18*4, the horsepower. 

393. Example. — A ball of 4 lbs., moving to the north on a 
smooth, level table with a velocity of 6 feet per second, strikes 
another ball, and after the collision is found to be moving at 5 
feet per second to the east. What was the amount of the 
impulse? If the collision lasted 0*002 second, what was the 
average force of the blow ? 

Ans. Subtracting the second velocity (in the way vectors 
are subtracted) from the first, we find that the sudden gain of 
velocity was ^ 61 feet per second in a direction making 
tan. -1-1 east of south. The impulse given to the ball was 

4 

therefore in this direction, and of the amount ^^q^ y61 pound- 
seconds. The force, therefore, was in this direction, and of 

4 — 
the average amount -^^ y 61 -r 0*002 lb., or 485 lbs. 

EXEECISES. 

1. Compare, the amoiints of momentum in a pillow of 20 lbs. which 
has fallen from a height of 1 foot, and an ounce bullet moving- at 200 feet 
per second. Ans., 12-8 to 1. 

2. A ball of 56 lbs. is projected with a velocity of 1,000 feet per 
second from a gun weighing 8 tons. What is the maximum velocity of 
recoil of the gunP Jns., 3-1 feet per second. 



494 APPLIED MECHANICS. 

3. There are two "bodies whose masses are in the ratio of 2 to 3, and 
their velocities in the ratio of 21 to 16. TMiat is the ratio of their 
-momenta ? If their momenta are due to constant forces acting on the 
bodies' respectives for times which are in the ratio of 3 to 4, what is the 
ratio of these forces ? Alt'!., 7 to 8 ; 7 to 6. 

4. A body weighing 10 lbs. impinges on a fixed plane with a velocity 
of 20 feet per second. If the coefficient of restitution e is 05, find the 
velocity of rebound, and how many foot-pounds of energy are wasted in 
the collision. Ans., 10 feet per second; 93-17. 

5. A ball of 6 ounces strikes a bat with a velocity of 10 feet per 
second, and returns with a velocity of 30 feet per second. If the duration 
of the blow be ^ second, find the average (time) force exerted by the 
striker. " Ans., 0-32 lbs. 

6. A body weighing 50 lbs. moving at the rate of 10 feet per second 
overtakes another body of 25 lbs. moving at the rate of 6 feet per second. 
If both masses be perfectly elastic, find their velocities after the shojk. 

Ans., 7^ and 11^ feet per second. 

7. A hammer-head of 2h lbs., moving with a velocity of 50 feet per 
second, is stopped in 0-001 second. TThat is the average force of the 
blow? ^«5., 3,882 lbs. 

8. A shell bursts into two fragments, whose weights are 12 and 20 lbs. 
The former travels onward with a velocity of 700 feet per second, and 
the latter with a velocity of 380 feet per second. What was the 
momentum of the shell when the explosion took place ? Ans., 496-8. 

9. A shell weighing 20 lb?, explodes when in motion with a velocity 
of 600 feet per second. At the moment of explosion one-'hird of the 
shell is reduced to rest. Find the momentum of the other two-thirds. 

Ans., 372-7. 

10. Water is flowing through a service pipe at the rate of 60 feet per 
second. If the water be brought to rest uniformly in one-tenth second 
by closing the stop-valve, what will be the increase of the pressure of the 
water near the valve, the pipe being taken as 50 feet long, the resistance 
of the pipe and the compressibility of the water being neglected ? 

Ans.jAOS lbs. per sq. in. 

394. As for the ^'ay in which the vibrations take place in 
two colliding bodies, if mathematically treated it is generally 
difficult : but very good working notions for the engineer 
are derivable from the results of experimental study, the 
subject being taken up in books on physics under the head 
" Acoustics." The best mathematical treatise is Lord Eay- 
leigh's " Sound." The time of an impact is known in several 
cases in which it has been calculated from theory, (vSee Art. 
404.) 

395. Let us consider what takes place when two ivory balls 
come together. There is a certain instant after they first 
touch when their centres move together just as if they were 
composed of soft clay — then they act on each other with their 
greatest pressure ; they are in their most strained condition, 



APPLIED MECHANICS. 495 

and supposing no loss by internal friction, the strain in the 
balls represents an amount of stored-up energy (see Arts. 259, 
267) equal to the kinetic energy which the bodies have lost. 
It is very important to remember this fact, that if bodies are 
to return to their old states after the collision, we must 
suppose that during the collision there is a storage of kinetic 
energy in the form of strain. All the kinetic energy will not 
be given out again, nor can we say that it is all stored, because 
there is a sort of internal friction causing part of the strain 
energy to be converted into heat when any change occurs. 
Now, if the whole of the stored-up energy is confined to one 
portion of the body, the strain may be too great. Thus, a 
steel rod 1 square inch in section, 1 foot long, will store up 
167 foot-pounds of strain energy in its stretched condition 
before it breaks. For suppose breaking stress to be 100,000 
lbs. per square inch. This will occur when there is a length- 
ening or shorteiiing of -0033 foot, so that the energy stored 
up is the work done by a force whose average amount is 
50,000 lbs. acting through -0033 foot, or 167 foot-pounds. If 
2 feet of the same rod stored up this same amount of energy, 
there would only be 83 foot-pounds in each foot of its length ; 
and it is easy to see that the stress is no longer the breaking 
stress of 100,000 lbs. per square inch, but only 70,700 lbs. 
per square inch. As we store the same amount of energy in 
smaller and smaller portions of a body, it is evident that we 
must approach a condition of fracture. 

396. We see, then, that at the place where contact occurs, 
two bodies, a and b, are strained ; but if A is of some very 
elastic material, such as tempered steel, the strain energy is 
conveyed very rapidly to every part of the body ; whereas if b 
is a feebly elastic body, the strain accumulates at one place, 
leaving the rest of the body unstrained, whilst at this place 
the strain may produce fracture. This slowness to communi- 
cate strain to the rest of the body may also be produced by 
the shape of the body. For instance, a rod struck sldewise or 
a thin plate struck in the middle does not so immediately com- 
municate its strain to the remote parts as a rod struck end- 
wise. Again, the nature of the parts of A and b in contact 
may be such that not only does the strain energy leave this 
part of A rapidly, but immediately in the neighbourhood of 
the place of contact there is a greater capacity to bear strain 
energy without rupture than is the case with b. Thus, when 



496 APPLIED MECHANICS. 



A rams the broadside of ship b, the side of b is bent 
inwards and the strain energy produced is accumulated near 
the place of contact till fracture occurs ; whereas, not only is 
a's stem able to transmit to all parts of A with great rapidity 
the strain energy which must be stored up in the whole mass, 
but at the stem itself the material of a is capable of with- 
standing greater stresses than the material of b's side. 
Suppose, however, that a's stem is not of steel, still b's iron 
or wooden side will be perforated if a has enough velocity ; 
a's stem may also be damaged in the impact in such a case. 

397. A candle may be fired, it is said, through a thin 
deal board with very little injury to its shape, and the usual 
explanation of this phenomenon given in books is that the 
candle has not time to get broken. This explanation is not 
patisfactory ; it is a little too vague. If we had the board in 
rapid motion, and striking the candle in the same relative 
position, the candle having previously been at rest, would the 
candle perforate the board 1 There cannot be any doubt that 
it would. Hence it is not the body struck which must in 
every case get hurt ; the pressure on one is equal to that on 
the other. Suppose ship a rushes at ship b when b is broad- 
side on, and rams her, b will probably be sunk, even if she is 
a much larger and better ship than A. But suppose that b is 
able to meet her adversary stem to stem, if they are equally 
strong they will equally injure one another, and if B is the 
stronger a will suffer the most. This case differs very much 
from that of the candle, because we can assume greater 
strength even for slowly applied pressures from stem to 
stern of the ship a than from side to side of B ; whereas the 
strength of the candle for slowly applied pressures cannot be 
compared with that of the wood which it punches from the 
board. What is meant by the usual explanation, " The candle 
has not time to get deformed " ? Why has not the soft candle 
time to get broken, and yet the wood has time to get torn 
asunder 1 The fact is, the wood, if it were slowly pressed, 
would communicate its strain energy to every part of the 
board and its supports ; but this communication takes an 
appreciable interval of time, however suddenly the pressure 
may be applied, or however great it may be. As the strain 
energy is rapidly produced it becomes accumulated near the 
place of contact to such an extent as to produce fracture of the 
wood. Now the point of the candle is subjected to the same 



APPLIED MECHANICS. 497 

pressure as the wood, and begins to get spoiled in shape — that 
is, it is compressed — and this compression produces a lateral 
spreading. In the meantime, however, the compressive strain 
energy is communicated very rapidly backwards along the 
candle, and the spreading and spoiling goes on along its entire 
length, but is small at any point, since it is distributed over 
the whole mass. Practically, therefore, the spoiling occurs 
only at the point of the candle, since time is needed for 
fracture of the material. 

398. An earthquake, when it acts on a house, usually 
tends to move it through a distance of probably a very small 
fraction of an inch, but it does this in a very short time — that 
is, the house gets a considerable velocity. The mass of the 
house multiplied by the greatest velocity, and divided by the 
short time during which the momentum is being communi- 
cated, gives the pressure which the foundations of the house 
are subjected to. Now, when the foundations are not very 
rigidly connected with the ground, the time of communication 
of the momentum is lengthened, and the pressure is conse- 
quently diminished. This is the usual Japanese plan of 
providing for earthquake effects. Unfortunately, the very 
means taken to diminish the pressure on the foundations also 
diminishes their capability of withstanding forces, and it has 
not yet been decided what sort of a house is best fitted to with- 
stand destructive earthquakes. Want of rigidity, combined 
with strength or toughness in the materials, and especially the 
quality of internal friction in the materials, so that vibrations 
may rapidly die away — these are the qualities needed. They 
are found in steel, wrought iron, and wood, and especially in 
wicker-work, in a less degree in cast iron and in brick or stone 
set in cement, and less still in brick and stone set in bad 
mortar. 

399. When you in some way understand the possibility of 
a candle perforating a board, you will be able to comprehend 
how sand, when blown in air against tempered steel, is able to 
abrade it ; how the emery wheel and grindstone going at great 
velocities are able to cut into hard metals ; and how in Cali- 
fornia a jet of water going with very great velocity is used for 
mining purposes instead of iron tools. 

400. Quasi-Rigidity Produced by Rapid Motion. — A top 
when not spinning can with difficulty be balanced on its point, 
and if left to itself it almost instantly falls ; whereas when it 



498 APPLlliD MECHANICS. 

is spinning the effect of slightly tilting it out of the perpen- 
dicular is not to make it fall, but to make it take a slow 
precessional motion. 

There is a piece of apparatus called a gyrostat, which is, 
in a more or less perfect form, to be found in every mechanical 
laboratory, and the student ought to experiment for himself 
with this apparatus on the curious effects of qua^i-rigidity 
which manifest themselves in tops and other spinning bodies. 
If he has a slight acquaintance with astronomy he will be 
interested in tracing the connection between the behaviour of 
a tilted top and the precession of the earth's equinoxes. 

When a circular sheet of drawing-paper is mounted like 
a very thin grindstone on an axis, and is gradually made to 
rotate rapidly, it is found to have become quite rigid — that is, 
it greatly resists bending as if it were made of steel. In the 
same way a long loop of rope, hanging round a high pulley, 
which gives it a quick motion, takes a certain form which it is 
very difficult to alter, as may be shown by striking it with the 
hand or with a stick : it resembles more a rigid rod than a 
flexible rope. 

Again, in the well-known lecture-experiments on smoke 
rings, we see that these little whirlpools of air have many 
properties in common with elastic solid bodies on account of 
the partial rigidity which is due to their rapid motion. 

In objects which spin and rub on a level surface, like tops, 
v/e have the interesting general rule, "Positions which are 
stable when there is no spin, are unstable when there is spin, 
and vice-i^ersd." Students of statics insist on a low centre of 
gravity in vehicles; students of dynamics sometimes insist on 
a high centre of gravity, as giving greater stability when there 
is rapid motion. 

It would be beyond the scope of a book like this to explain 
these curious phenomena, and I merely direct the attention of 
students to these instances in order to incite them to make 
experiments, and to seek for the explanation of what they 
observe. My popular lecture on Spinning Tops may be worth 
reading. 

401. Motion Produced by a Blow. — When a body sub- 
jected to a blow is quite free to move in any way, unless the 
blow acts through its centre of gravity, the body will not 
merely move as a whole, but it will revolve. When the blow 
acts in a direction through the centre of gravity there is no 



APPLIED MECHANICS. 



499 



K,- 



rotation produced. It is usual in such a case to consider the 
motion of the centre of gravity of the body, and the motion of 
the body about an axis throngli its centre of gravity, for it is 
known that any motion whatsoever 
of the body consists of a combination 
of two such motions. It is found v. . 

that the kinetic energy communicated 
to a body by means of a blow i? 
best calculated in the following way 
if we know the nature of the motion . 
— First, find the kinetic energy, a* 
if every portion of the body had the 
motion of the centre of gravity. 
Secondly, find the kinetic energy of 
rotation as if the axis of rotation 
through the centre of gravity were 
fixed in space. Add these two re- 
sults together. We may regard from 
another point of view the instan- 
taneous motion of a body when it is 
struck — namely, as a rotation about 
some axis which does not itself move. 
This is only the case for an instant ; 
immediately afterwards we must re- 
gard the body as moving about another 
fixed axis. If a body is hinged so 
that it can only move about a fixed 
axis, it is always possible to find the 
point at which the body may be 
struck, and the direction of the blow, 
which will tend to produce an instan- 
taneous rotation about this particular 
axis, and therefore to produce no 
pressure at the hinge. Thus the 
ballistic pendulum of Fig. 278 is _ 

always struck in such a way, and Pig. 278. 

the point in which it is struck is 

called the " centre of percussion." An easy way to find the 
centre of percussion is as follows : — Make the body vibrate 
like a pendulum, about its axis of suspension, under the action 
of gravity. Now find the length of the equivalent simple pen- 
dulum. This is the distance of the centre of percussion from 




500 APPLIED MECHANICS. 

the axis. In a tilt hammer all blows ought lo be delivered 
from this centre of percussion if we wish to have no pressure 
on the bearings. A cricket-bat or a rod of iron tingles the 
hand when we strike a blow with it, nnless we happen to 
strike afc the centre of percussion. For a rod of iron free to 
move about one end, the centre of percussion is at two-thirds 
of the wav towards the other end, (See Art. 454.) 

402. The Ballistic Pendulum of Fig. 278 is a contrivance 
which enables us to measure the velocity of a bullet. It con- 
sists of a mass of wood, a, forming part of a pendulum. The 
bullet is fired into it, and the wood swings backwards in con- 
sequence. The ballet is fired in such a way that it will 
cause no jar to he given to the pivot b. The momentum 
existing in the bullet before it enters the wood belongs now to 
the whole mass of which it becomes a part, c is a silk ribbon 
which is pulled through a moderately tight hole or over the 
edge of a table by the swing of the pendulum, and the length 
of ribbon pulled through is found to be proportional to the 
momentum of the bullet before entering the wood. 

If i^ is the weight of the bullet, its horizontal velocity in the 

plane of s^^vinging "being v, its momentum is — r. The moment of 

momentiuii z - v {H x ia the perpendicnlar distance from b to the 

axis of the bullet) before impact is equal to the moment of 

momentom after impact ; and if i is the moment of inertia of a 
and the hnllet about b after collision, and a is its initial angular 

velocity, then i a = x — v . . . . {!). A certain part of the whole 

kinetic energy has been converted into heat ; the mechanical 
energy now in the system, 1 1 a-, will he converted into potential 
energy, w being the total -weight, the centre of gravity wiU he 
raised through the distance h such that w^ = ^la- . . . . (2) ; so 
that, as a is kno^wTi, h is known. KJiowing h, we can either find d, 
the total swing, or we can calculate or find experimentally the 
length of the rihbon which may be drawa up, and from the length 
of the ribbon we can calculate r. 

In the short time when the bullet is being lodged it is exercising 
horizontal force at each instant, and there may he horizontal force 
at B in the same direction. AVc may say that these external forces, 
acting at the centre of gi-a-aty, are balanced hy the mass of the 
whole body miiltiplied by the acceleration of the centre of gravity 
at each instant, and hence their integi-al effects balance ; that is, 

then, the impulse — c 4- p (where p is the horizontal impulse from 



APPLIED MECHANICS. 501 

W 
the point of support in tlie same direction) = - « . b g if g is 

the centre of gTavity, because a . bg is the velocity produced in 
G, and w is the weight of the pendulum and bullet Hence 

p= - a . liG - - V . , . . (3), 
9 9 ' 

This is 0, therefore, if 

AV . a . B G — IVV .... (4), 

But by (1), (4) means that w . a b g = ^' Jc^a/x, or a; = F/b g. 
Now this is the rule (Art. 454) by which we find the length 
of the equivalent simple pendulum or the distance to the jDoint 
p of oscillation (Art. 454). It also for this reason gets the name 
" poiat of percussion." 



APPENDIX TO CHAPTER XXIII. 

403. In the Example, Art. 385a, we had an example of stop- 
page of water in a pipe as if all the water had exactly the same 
motion. It was an interesting academic exercise. The following- 
exercise is also interesting, and will give us more information. 

Sudden Stoppage of Water in a Pipe of Uniform Section. — 

Suppose the pipe to be infinitely rigid. It will be found that the 
effects are independent of whether the pipe is vertical or horizontal, 
but we shall consider it to be vertical. It will be found on closer 
examination that if we know the pressure we need only con- 
sider, in our problem, p, the pressure above the normal pressure, 
and this is what we shall do. Let v be the axial velocity 
of a particle of water, w the weight per unit volume, so that 
wig is its mass per unit volume, often called p; if k is the 
cubical elasticity (during quick changes of volume) and t 
is time ; if a: is distance to a point q measured along the pipe in 
the direction of motion, and if the pipe is taken of unit cross- 
sectional area; if x is the total distance already travelled by 

a particle at q, so that at a; + 1, x becomes x + —, fluid 

dx 

which once occupied length 1 of the pipe now occupies length 
1 + -^j so that its increased volume is - — . Hence, as pressure- 
producing volum.etric strain = k. x compressive strain, 

f= -K^....(l). 

Now velocity v = -rr, and as the mass between x and x + Sx is 
dt 

p . 5x and the pressures are p and p + 5p, we have the force - Sp 



502 APPLIED MECHANICS. 

causing acceleration -j-^ in the mass p . Sz, and hence 

or, from (1), k -^ = p —^, . If K/p or Kg/w be called a^, we have 

and we recognise this as the equation of wave propagation with 
the velocity a. Differentiate (3) with regard to t and use v for 

ds-ldt, and we have a--^^ = vs. 
' dx^ dt^ 

Let OUT prohlem be this. When water is flowing with a 
"uniform velocity - Vq, let an infinitely rigid, thin diaphragm 
suddenly produce a stoppage at a; = ; what will happen ? Con- 
sider the pipe only for positive values of z. "We have v = - v^ 
everywhere at ^ = 0, and at a; = we have v such a function of 
the time that it was - t\ 
is 0. 

To make things simpler, let v^ be a new variable such that 
t'l = V + f,,, and therefore the conditions are that v^ = every- 
where at ^ = 0, and at a: = we have r^ such a function of the time 
that it was till ^ = 0. and ever afterwards v^ = Vq. "We see that 

'-d^ = d^-- ■^'^' 

If we let J- be denoted by 9 in the usual symbolical way, (4) is 
dt 

SolAring this as the simplest linear equation would be solved if 6 
were a quantity independent of x, we have 

if A and b are functions of the time corresponding to a; = 0. As 

X may be infinite, b = ; so that 

,.i^,-xt*/ay(^) .... (5), 

where / {t) is till t = 0^ and tlien is v^. and remains of the 
value i-'o ; so that 

^' = - t'o +fii - ^M (6).* 

"We see now what occurs. Until t - x/a = or t = x/a, v = - Vq 

at any place ; and after t = x/a, ?• = 0. Hence we have from 

= a state of no velocity spreading along the pipe with the 

♦ The symbolic e^^/(i) means /(« + 6) 



APPLIED MECHANICS. 503 

velocity a to infinity. Since v = — , if we integrate (6) in regard 

to time, we have 

X = - v^t + ^ {t - x\a) .... (7), 

where f is such that —^ — = / (t/), whatftver y may he. Hence, 

ay 

using (l),i? = — /(^ - xla)\ or, since 

we have p — \/ — -/(^ - ^/^) • • • • (8)- 

This tells us that^ = till zJ - xja = 0, and then is Vq a/ — > ^^^ 

remains of this value. This is the state of pressure produced every- 
where with perfect rest accompanying it, at the velocity a from the 
place of the sudden stoppage if the pipe is infinitely rigid. 

If at a; = ^ there is a place where the pressure is kept constant 
^^we say p = there), at zero, our wave is refiected as a wave of 
velocity + Vq and no pressure, till on reaching a; = it is again 
refiected as a wave of no velocity, and so on. As the pipe is not 
infinitely rigid, and as there is friction, the wave really diminishes 
in its values of velocity and pressure as it travels, but we may take 
a pressure ajjproachirig the value Vq \/-K.wlg as being instantly pro- 
duced by a sudden stoppage. Experiments with suddenly closed 
valves give measured pressures in the laboratory considerably less, 
partly because the stoppage is not one of infinite suddenness, partly 
because of the inertia of the pressui-e-measuring apparatus. In the 
case of water the pressure in pounds per square inch is 20 Vq if Vq is 
the velocity in feet per second. 

This great pressure produced by stoppage is taken advantage of 
in the hydraulic ram. If the bottom of the sea were smooth, and 
a sea with a translational velocity Vq were suddenly stopped on its 
motion by a vertical wall, the pressure on the wall would be what 
we have given above, for a very short time, if the wall were per- 
fectly rigid ; what would occur subsequently I do not know, be- 
cause there is atmospheric pressure at the surface of the sea. We 
know enough, however, to see the necessity for some springiness at 
the wall surface. It is for the same sort of reason that heavy seas 
produce so much damage sometimes when objects are struck by 
them on board ship. Curious stories are told by sailors of half- inch 
bars of iron being bent and broken by seas coming over the bul- 
warks, and it is just possible that they may be true, although it is 
more probable that such fractures are due to blows from passing 
wreckage. Anyone who has seen the ruined breakwater at Wick 
will believe in the greatness of the forces due to blows from ocean 



404. Students will find it an excellent easy mathematical exercise 
to assume that in an infinite length of pipe filled with water there is 
a piston whose displacement is a pure function of the time. The 



504 APPLIED MECHANICS. 

problem is identical with the simplest problem in Telephonic 
Signalling. If the pipe is supposed to yield and to be leaky, and 
if the viscosity of the liquid is considered, we have the same 
problem as that of the Fhilosophical Magazine, page 223, August, 
1893. 

When a prismatic bar of length l-y and velocity v-^ in the direc- 
tion of the common axis overtakes a bar of length l^ (greater than 
J;), moving with a velocity v^, the bars being of the same material 
and cross section, jMt. Love says that the ends at the junction 
move with a common velocity | [v-^ + v^, and a compressive strain 
i ("^1 - ^'a)/^^ is produced, a is the velocity of a longitudinal wave 
of sound, or e = a^p where e is Young's modulus and p is the mass 
of the bai- per unit volume, or our w/g above. Waves of compres- 
sion rim fi'om the junction along both bars, and each element of 
either bar, as the wave passes over it, takes suddenly the velocity 
i (^'i + "^s) JWid the compression ^ (^-i - v^^ja. 

When the wave reaches the free end of the shorter bar it is 
reflected as a wave of extension ; each element of the bar as the 
wave passes over it takes suddenly the A^elocity which initially 
belonged to the longer bar and zero extension. After a time, equal 
to twice that required by a wave of compression to travel over the 
shorter bar, this bar has uniform velocity, equal to that which 
originally belonged to the longer bar, and no strain. 

The impact now ceases, and there is in the longer bar a wave of 
compression of length equal to twice that of the shorter bar. The 
wave at this instant leaves the junction, and the junction end of 
the longer bar takes a velocity equal to that which it had before 
the impact. The ends, therefore, remain in contact without pres- 
sui-e. This state of things continues until the wave returns 
reflected from the fui'ther end of the longer bar. When the time 
from the beginning of the impact is equal to twice that required by 
a wave of compression to travel over the longer bar, the junction 
end of the latter suddenly acquires a velocity equal to that origin- 
ally possessed by the shorter bar and the bars separate. The 
shorter bar rebounds without strain and with the velocity of the 
longer, and the longer bar rebounds vibrating. 



505 



CHAPTER XXIV. 

FLUIDS IN MOTION. 



405. We tried in Art. 145 to give exact notions on the 
subject of pressure in fluids not in motion. When we 
supposed that the weight of the fluid itself was insignificant, 
we found that the pressure on each square inch of surface 
touched by the water, and on each square inch of interface 
separating two portions of the fluid, was everywhere the same. 

One of the best methods of observing the pressure, any- 
where, is by inserting a pressure gauge. The gauge indicates 
the pressure per square inch at the part of the liquid to which 
a communicating little gauge tube penetrates. It always does 
so when there is no motion of the fluid at the point in ques- 
tion ; but unfortunately, when there is motion there, the 
introduction of the tube alters the motion there and more 
or less falsifies our measurement. 

We are now about to consider pressure in fluids in motion, 
and we approach our subject by first speaking of pumps. 

406. A pump is a machine which gives energy to water — 
that is, it can raise water to a height, giving it potential 
energy. It can force water into a vessel under great pressure, 
moving pistons — that is, it can give to water pressure energy. 
It can set water in motion — that is, give it kinetic energy. 

Reciprocating-pumps are either lift-pumps, or force-pumps, 
or combinations of both. 

In Art. 137 we described the force-pump used with hydrau 
lie presses. In principle this is the same as all other force- 
pumps. The feed and other pumps of steam-engines, and the 
pumping parts of pumping engines, are described in detail in 
works on those subjects. As to Lifting Pumps, in Fig. 279 
we have a diagrammatic representation of the common village 
or house pump. The rod k, usually worked by means of a 
handle or lever, pulls the bucket, b, up and down. The 
bucket has openings arranged with flaps or other valves in 
various ways, so that fluid may pass upwards but not down- 
wards. At w there is another valve w^hicli opens upwards. 
First suppose there is only air between b and the well, j. 
Sometimes in the morning it is necessary to throw in some 



506 



HAVIC^ 



water 
mnct 



JW\ 



: I _ ake it air-tightu As B desc@Dtds, 
Old w passes tfarongfa r As b is 
irr^i 11 produces a partial ▼acumn below 
it ; air passes firom h to the barrel, »id 
water rises in h. As the pDmping pro- 
ceeds there is more of a Tacaam until 
wat^* fills H and the barrel, in case b is 
not- nearly 34 fe^ above the lerel in the 
well 25 feet is probably the maximrai! 
hoght in house pumpsL After this it is 
water that passes fhrcma^ B in ever j down 
sfanoke, and is lifted in every up sfarok& 

Whoi a hi^ber lift is needed, we 
sometimes place a valve in the npper part 
of the barrel, or in ilie delivery {Hpe, to 






As more and more war, 



prevent the rrnr: 
Kg. 280 is a :.L^- 
grammatic repre- 
sentation of a 
foree - pump. 
Sometimes apiston 
is nsed instead of 
the plunger, p. A 
is an aiT-vesseL 
nters at B, 
lines great 
^7 up the 
case the 
Un- 
solve 



tir !Ift-^ WAter. 



deKvery pipe, z Iii 

stream is not - : tI 

fortunatdy tiie ?-i: 

in the water &ji _ t \~ \-- :■- 

rate whidi dei : i. : ~ _ 

air is already iii liic miSJts:. jfig. 'J^\ 

shows a double-aetiiig foree-piimp. 

When tJie pistrai b moves to tlie ri^t, 

water passes tiirough the valves, A, 

to the delivery pipe, d, and ent^s ^e ¥%. saa 

barrel iJiron^ d from the sndian 

pq)e, s, and well Whoi B moves to the Irft, c is tiie deKvery 

i^lve, and E is the suction valve. ESg. 282 shows a ccMnbined 

force and lift pump whose action is very eaaj to understand. 

The deliverv valve is net shewn. 




APPLIED MECHANICS. 



507 




407. A pump must be efficient — that is, it must do nearly 
much work on water as the pump itself receives from an 
engine or labourers. But it must be 
remembered that the best pumps used 
for different purposes have very differ- 
ent efficiencies. Forty per cent, would 
be regarded as a reasonable efficiency 
for reciprocating pumps of low lift, 




S I Fig. 281. 

whereas it would be regarded as rather 
_ ^^_ poor for pumps of high lift. 

Example. — (1) Suppose that when a 
pump is deliTering- a certain quantity 
Q = 1,000 gallons per minute, there is 
a loss of 35 foot-pounds per pound of 
water in the pump and horizontal pipes> 
and 0-05 foot-pound per poirad for each foot of vertical pipe. 
Calculate the total loss per pound in each of the following cases, 
the vertical height of pipes and of delivery heing called h feet : — 





Useful work 








h 


per pound of 


Lost energy. 


Total energy. 


Efficiency. 




water. 








20 


20 


36 


56 


0-36 


50 


50 


37-5 


87-5 


0-57 


100 


100 


40 


140 


0-71 


200 


200 


45 


245 


0-82 


300 


300 


50 


350 


0-86 


400 


400 


55 


455 


0-88 



(2) If the loss of the pump is 20 + 15 x 10"^ q per pound of 
water, and of the vertical pipe is 5 x 10~^ q^ per pound, q heiag 



508 



APPLIED MECHANICS. 



gallons per minute, work out two more tables like the above — 
one when the delivery is 500, and the other when it is 1,500 gallons 
per minute. 

408. The peculiarity of reciprocating-piimps is that, when 
there is no slip, there is the same quantity of water passedthrough 
the pump at every stroke. If we 
know the size of everything, then 
the speed of the pump tells how 
much water is delivered, and if we 
know the height to which it is 
delivered, we know the work done. 
Now, in a centrifugal - pump 
things are somewhat different ; with 
a given speed of pump we may 
have very different quantities of 
water passing. We may have the 
pump running at a certain speed, 
and no water being delivered, and 
very little work being done, only 
frictioual work, in fact. Now a 
slight increase of speed may cause 
an abundant flow of water, and a 
tremendous increase in the work 
usefully done. Strange to say, if 
this speed be now diminished to what 
it was at first, it does not follow 
that the water will cease to flow. 
In any centrifugal-pump such as 
Fig. 28.3 we observe that there is a 
central wheel, A b, with vanes, which 
can be rotated very rapidly. Water 
can enter the wheel on both sides, 
c C, at its centre from two supply 
pipes, D D, which meet in one pipe, 
E, below. The water fills the wheel 
or space between the vanes, and, being 
whirled round with great velocity, 
tries to get away at the circumfer- 
ence of the wheel, because of the cen- 
trifugal force, and it flows out into the casing, f f, which 
gradually becomes the discharge-pipe of the pump, as shown 
in the smaU scale drawing, Fig. 284. This is a popular 





Fig. 284. 



APPLIED MECHANICS. 509 

explanation of what occurs, but we must examine the matter 
more carefully. 

409. What occurred when the pressure in the pump of the 
hydraulic press became greater than the pressure of fluid in 
the press ? There occurred a flow of fluid. The fluid was set 
in motion from pump to press. A difference of pressure 
between two places which communicate with one another 
usually means a tendency to produce motion. 

In the hydraulic press the flow is intermittent. Why? 
Because the pressure is intermittent. We may be sure that 
when we have a certain difference of pressure between two 
places, and this is always the same, the flow of fluid is per- 
fectly steady ; and we saw in Art. 408 that work is then done 
on the fluid with perfect uniformity. We also saw how to tell 
what work was done. The work done on a fluid in a minute 
is simply the difference of pressure per square foot which causes 
the flow, multiplied into the number of cubic feet of water 
which flow per minute. This gives the answer in foot-pounds. 
Suppose that with pumps, or in any other way, we establish 
a difference of pressure between a place A and a place b, and 
suppose we know that our pumps and other arrangements will 
not break down — in fact, that the difference of pressure 
between A and b is really a fixed thing on which we can 
depend — we know that the flow of water from a to b will be 
the same at all times, and that the same amount of work will 
be done upon it every minute. If, now, we leave out of our 
minds all consideration of how that constant difference of 
pressure has been produced — merely think of the two vessels 
A and B — we know that this difl'erence of pressure which has 
been established is really a store of energy. What enables us 
to call it a store of energy ? The fact that w^e know it will 
not be suddenly destroyed. 

Suppose we know that a man has a certain income paid, 
say, by Government, and suppose we are perfectly certain that 
this income is constant, we can regard the certainty of the 
man's income as a store. To say that a man makes a sovereign 
in a day is not of much importance, but to say that the man 
has a regular income of one pound a day makes him a respect- 
able member of society, and a store of social energy. 

A man may be sitting in Parliament, but this in itself does 
not make him a store of political energy ; whereas if we know 
that he is certain to sit there for a length of time — that a 



510 APPLIED MECHANICS. 

general election or general vote of the House is unlikely to 
unseat him — we can regard him as possessing a store of 
political energy. 

Similarly^ a pound of water in the vessel a, at rest, 
possesses more energy than a pound or water in the vessel 
B, at rest. (We must remember that there are some means of 
keeping the pressures in a and b what they were.) How much 
more energy has it ] If the difference of pressure is p lbs. per 
square inch, then it has 2*3 p foot-pounds of energy ; not in 
virtue of its own intrinsic worth, but because it is where it is, 
and because we know that if it flows into the vessel b, nothing 
'will alter in the pressure conditions of the two vessels till it 
gets into the vessel B. We understand, then, that a pound 
of water, subjected to a pressure of p lbs. per square inch, may 
be said to have a store of 2*3 p foot-pounds of energy, if we 
know that the motion which is occurring in the water is steady, 
and is not altering capriciously. 

Thus a pound of water at the pressure of the atmosphere, 
14*73 pounds per square inch, possesses, in virtue of this pres 
sure, 2 "3 X 14 "73 or 34 foot-pounds of energy. It would 
possess the same energy if it were at the pressure o, the 
pressure in a vacuum, but were 34 feet higher than it is in 
position. 

Suppose that a is a closed box, and that it is filled with 
water, under great pressure. Now, suppose we open a valve, 
and let this wat^r escape. Although there was a great pres- 
sure, p, just for an instant, and therefore a rapid flow of water 
just for an instant, this almost instantly dies away, because 
the pressure in a is almost instantly diminished. Every 
pound of the water did not then have a store of 2-3 p foot- 
pounds of energy, and yet it was at the pressure of p pounds 
per square inch. It is the certainty that the state of pressure 
in A will continue constant that gives to pressure its signifi- 
cance, and gives to us the liberty of regarding pressure as a 
store of energy. 

410. Suppose that we have, anywhere, steady motion of 
water. Consider a pound of the water. What is its total 
store of energy ? 

1. It is A feet above some datum level. Then if one 
pound of water ever were allowed to fall to the datum level, 
it wotild do h foot-pounds of work in falling. The mechanical 
energy stored up in a miller's dam is simply the weight of thf 



APPLIED MECHANICS. 



511 



water, multiplied by the height through which it can fall. Of 
course, if any other volumetric force acts on the pound of 
water, this will constitute another store of potential energy. 
We are supposing that the weight of the w^ater is the only 
volumetric force. 

2. As the motion is steady, if the water is at a place 
where pressure is j^ pounds per square inch, it possesses, in 
virtue of the fact that the motion is of a steady character, the 
store 2-3 p foot-pounds of energy. 

3. As the water is in motion, if v is its velocity in feet 
per second, as its mass (the mass of one pound) is l-i-32-2^ we 
know that its kinetic energy, or energy of motion, is the 
square of the velocity divided by 64-4. 

Exercise. — Calculate the numbers in the following table, 
which shows the relative values of h and p and -y, if we wish 
to convert one of these forms of energy into another. Thus, 
bhe energy due to a difference of level of 2*3 feet is equiva- 
lent to that due to a difference of pressure of 1 lb. per square 
inch, or to that due to a velocity of 12-18 feet per second. 



Difference of level. 



Pressure. 



Velocity. 



2-3 feet. 
16-1 „ 
34 „ 
64-4 „ 



1 lb. per square in. 

• j> >) 

14-73 „ „ 

28 



12-18 feet per second 

32-2 

45-7 

64-4 



411. Now, we shall not suppose that the pound of water has 
any other stores of energy than these. We know that, as it 
is compressible to some extent, it may have a store as the 
mainspring of a clock has. This store must always be taken 
into account when the fluid is air or any other gas. It may 
also be electrified or at a high temperature, or it may have 
other stores of energy which we are neglecting. Merely think 
of these three stores : — Potential energy due to height above 
a datum level ; pressure energy due to the unchangeableness 
of things ; kinetic energy due to its actual motion. 

What we must remember carefully is the fact that this 
pound of water retains all of this energy except what it 
loses in friction. Suppose that a man has capital in the 
shape of gold, capital in the shape of shares which never alter 



512 APPLIED MECHANICS. 

in price, and capital in the shape of a pet manufactory, which 
wastes money just in proportion to the amount of capital 
invested in it. He may buy more shares or sell tliem out, in- 
vest more or less money in the pet manufactory, but all the 
time his only loss is the loss from the manufactory. He may 
have no gold, or no shares, or very little money in the factory, 
but all the time his total capital is unchanging, except that he 
loses in proportion to the value of his factory. It is only when 
water has part of its energy in the shape of kinetic energy, only 
when it is in motion, that it loses any part of its total store. 

412. Consider a lake of water at rest. Consider a point A, 
and somewhat below its level another point b. A pound of water 
at A has the same store of energy as if it were 
\^ A -—p ^ at B. In neither case is there any energy of 
motion. The store of energy at A is merely due 
to height above some datum and the pressure 
Eig. ^85. pgj, square inch at a. Xow, if a pound of water 
f^ets to A from B, it loses potential energy li foot-pounds, if the 
difference of the level is A feet, and it ought to gain an equiva- 
lent of pressure energy, and the gain of pressure is, as we have 

already seen, simply ^ pounds per square inch. Thus, if A is 

34 feet, and jj is the gain of pressure, then there is a gain of 
pressure energv of 34 fool^pounds — that is. there is a pressure 
at B of 34-^2•3, or 14-73 lbs. per square inch greater than the 
pressure at A. This is an increase of pressure called one 
atmosphere. In still water there is an increase of pressure of 

one atmosphere for every 34 feet of 

^__^^_^* _j' , descent. (See Art. 173.) 

413. To further familiarise us with 

the idea, consider the flow of water from 

an orifice. 

In Fig. 286 we see the stream lines 

along which water flows out of the orifice. 

The shapes of these stream lines will depend 

on the shape and position of the orifice. 
Xow, a pound of water at the upper 
still surface a is at atmospheric pressure, and when it reaches 
c it is also at atmospheric pressure, so that its pressure 
energy remains the same. 

But at c it has fallen h feet ; it has lost A foot-pounds of 
potential energy, and it must therefore have gained 4 foot-pounds 




Ai»PLlEt> MECaANtCg. 



513 



c c 



c 



Fig. 286. 



of kinetic energy. If v is its velocity, then v^ _t. 54.4 must be 
equal to A, so that v may be calculated; hence its velocity is just 
the same as that of a stone which had fallen freely from A to c. 

To illustrate this, a high vessel may be used, from which at 
different levels tubes come out, ending in nozzleSj throwing 
jets vertically upwards. These 
jets do not rise to the same height, 
because there is a loss of energy 
due to friction, and this friction 
occurs principally at the nozzles. 
If the jet reaches within a dis- 
tance /ij of the level of still water 
inside, and if the nozzle is at the 
depth li below still water level, 
then if we may say that all the 
friction occurs at the nozzles, 
Aj-f-A expresses the loss as a frac- 
tion of the whole kinetic energy at the nozzle. 

When the measurements and calculations are made for 
different levels of the water in the vessel, we find nearly the 
same result in every case, showing that the friction al loss of 
energy seems to be proportional to the kinetic energy there. 

There is very little friction in the case we are considering 
in Fig. 286, where the orifice is sharp-edged, and we have a 
very simple statement of the velocity at c. 

Can we say the same about the velocity at b % Gertninly 
not. If we knew the pressure energy at b, we could say how 
much of the lost potential energy has been invested in this 
shape, and therefore how much has been invested in the 
shape of kinetic energy ; but without knowing the pressure at 
B, we cannot tell what is the velocity there. 

414. In this subject of flowing water there are more mislead- 
ing hypotheses, due to perverted ingenuity, than in almost any 
other ; and, unfortunately, the logical conclusions drawn from 
these hypotheses, when known to be untrue, are said to be the 
statements of theory as opposed to loractice. We often hear 
the statement, "the theoretical velocity at an orifice is the 
velocity which the water would have acquired if it had fallen 
freely, as in a vacuum, from still water level ; " whereas it is 
evident that we cannot tell the velocity at any point in the jet 
unless we know the pressure there, and we only know the 
pressure on the very outside of the jet. 



514 APPLIED MECHANICS. 

Now, although we do not know the pressure or velocity at 
every point of water flowing from an orifice, the studies of 
Professor James Thomson enable us to make certain im- 
portant stat-ements which agree with experiment. One of 
these is tliis : — 

When frictionless liquid flows from two similar vessels 
tlu'ough similar orifices similarly situated with regard to free 
water le^•el, the lines of flow are exactly of similar shape ; the 
velocities at similar points are exactly as the square roots of 
the dimensions of the vessels, and the total quantities of liquid 
which flow are proportional to the square roots of the fifth 
powers of the dimensions. 

Thus, if we have water flowing similarly from thi'ee similar 
vessels, all made from the same drawings, but to different 
scales — say one 1 foot, another 4 feet, and another 9 feet deep 
— the velocities at similar points are as 1 to 2 to 3, the 
sections of stream tubes are as 1 to 16 to 81, and the 
quantities of liquid flowing from the three vessels are as 1 
to 32 to 243 ; thus we are quite sure that 243 times as much 
liquid flows from the third vessel as from the first. 

Hence, suppose we want to know how much water is flow- 
ing in a small stream^ we dam the water up somewhere, and 
let it flow out of our dam through a notch like Fig. 287 (a 

_ _ ^-_- ^ right-angled isosceles notch) in a 

^1^^^_i:^^[ .li^^^t^ ') wooden board with sharp edge.s. 
p^,-^<l ^ I jj^jggjj^ ^g prefer to have the 

/ v^l ^ i edges of the notch made of metal, 

J^:'^ yf ^x :^ I so that we shall be sure that 

y^— '-- '--~^\ ^ \ I they are straight and sharp. 
^ v" ' Measure the height of the still 

Fig. 287. water in the dam above the 

lowest point of the notch (d). 
A graduated post, rising from the bottom of the dam 
some feet away, its zero being on the level of d, is very 
convenient for this. This one measurement tells us how 
much wat-er is tumbling over. For we know that, sup- 
pose there is rather a drought one day, and a shows the 
appearance of the notch ; and on another, a rainy day, a shows 
its appearance, we observe that the orifices through which the 
water is flowing are similar and similarly situated with regard 
to the still water level : and Thomson's theory enables us to 
say that, if on one day the height is 1 foot, and on another it 



APPLIED MECHANICS. 515 

is 4 feet, then 32 times as much water comes over on the second 
day as on the first. The flow of water is exactly proportional 
to the square root of the fifth power of the height d a. 

Now, Thomson measured very accurately how much water 
is flowing when the height is 1 foot, and he found it to be 2*635 
cubic feet per second. Hence we have the rule : measure the 
vertical height d a, at any instant, in feet ; raise this to the 
fifth power, and extract the square root, and multiply by 
2-635, and we know how much water is flowing in cubic feet 
per second. 

If we know the cubic feet of water flowing per second, we 
know the Aveight of the water, since a cubic foot of water 
weighs 62 "3 lbs., and the weight of water, multiplied by the 
number of feet through which we can let it fall, tells us the 
foot-pounds ])er second — the available power of the stream. 
The foot-pounds per minute, divided by 33,000, is the horse- 
power of the stream. 

415. Gauge notch observations, made from day to day on 
a stream, enable a person to make very exact calculations as 
to the power available for the driving of mills by turbines, for 
the working of hoists, cranes, and lifts, and for hundreds of 
other purposes. 

How Thomson used his theory in proving that the famous 
Lowell empirical formula, for rectangular gauge notches, is 
really a rational one, will be found in Art. 436. Students 
ought to treasure anything published by Thomson on fluid 
motion. 

416. Many rather abstruse-looking questions are easily an- 
swerable when we fully grasp the significance of the energy law. 

The fundamental fact which makes any hydraulic problem 
clear to you is this. If we may neglect friction, then a pound 
of water at any place has its total energy in three shapes. It 
has h foot-pounds of energy, because it is A feet above a datum 
level. It has 2*3 j; foot-pounds of energy, because its pressure 
is 2^ pounds per square inch ; and it has v^ -^ 64-4 foot-pounds 
of energy, because its velocity is v feet per second. 

To take another example : — 

Suppose we have a pipe which is in the main horizontal, 
so that we may neglect differences of level. Then we have to 
remember that the pressure energy, plus the kinetic energy, of 
a pound of water does not alter. When water flows along a 
pipe which is full, there must be the same quantity flowing 



516 



APPLIED MECHANICS. 



everywhere. We are sure, thereforej that there must be 
greater velocity wherever the pipe is contracted. But greater 
velocity meaiis greater kiiietic energy, and if this water invests 




more of its energy kineeiealiy, it must have less in the shape 
of preg.5=Jiire energy. That is^ the pressure of the water at b 
(Fig. 288) is less than at a or at c. The pressure at B may 
become very small indeed. It is easy, in this way^ to reduce 
the pressure to much less than the atmospheric pressure — 
merely contracting the cross section of the pipe is sufficient. 
We cannot make the pressure as small as that of a vacmrm, 
for before that limit is reached vapour forms. 

417. Suppose a conduit of this kind were carried over our 
fields, and that a quantity of water lay in our fields, at not too 
great a depth below the conduit. If we bring a pipe to the point 
B from the field-water, this becomes a suction-pipe, and we get 
our fields drained at the expense of the conduit owners. We 
spoil their water, if it is clean, but at aU events we get our 
fields drained. 

In this Kes the theory of jet pumps, and much of the 
theory of injectors, eta The jet pump of Professor James 




Pig. 289. 

Thomson simply consists of a small pipe, A (Fig. 289), which 
ends in a nozzle. Through tfais» let us suppose, we have a 



APPLIED MECHANICS. 517 

small supply of water flowing from some pretty high reservoir. 
Suppose that this water flows into the atmosphere at c. 
Evidently the pressure at a is much less than the pressure at 
c. It is less then than the atmospheric pressure, and hence 
the neighbourhood of « is a partially vacuous space, so that 
the pipe b e becomes a suction-pipe, and water tends to flow 
from a point at e, to b, and on to c. Thus, if we have a small 
supply of water from a high reservoir, we are able to drain a 
marsh with it. 

418. There is not space here to speak of the hundred other 
ways in which our principle comes in to simplify all sorts of 
puzzling phenomena. We may refer to Mr. James Perry's 
siphon for the discharge of flood- waters at the weirs in rivers. It 
has no moving parts in the water, being simply a wide, open pipe 
of varying sectional area, through which an object as large as a 
bullock might pass, without injury to the sluice. It has been 
found, by actual trial, that the quantity of water passing per 
minute through such a sluice is independent of the fall, so 
long as there is suflicient fall to balance the waste by friction 
in the pipe, a few inches being enough in the case of wide 
pipes ; and a velocity of 45 feet per second, at the smallest 
section, may be calculated upon when the sluice is working 
full power. The cross-section of the siphon is like the letter 
D at the tail ; the masonry, or concrete basin, is actually a 
portion of the siphon, and the horse-shoe shaped space between 
the lip of the basin and the iron edge of the siphon proper is 
the actual opening. The quantity of water passing at any 
time is regulated by the admission of air to, or its exclusion 
from the siphon. There is a throttle-valve arrangement by 
means of which the siphon may be adjusted at any time, so as 
to vary what may be called the normal water level in the 
reach above the sluice, and to vary the opening through which 
a constant stream of water falls on a ridge-shaped portion of 
the siphon. This stream of water is needed to exhaust the 
siphon of ail*, so that action may be set up at any time, even 
when the water is not passing over the top of the throttle- 
valve. The manipulation of such sluices is perfectly simple — • 
they may be made self-acting ; but it is proposed in important 
places to work all the sluices on a river from a single station 
electrically. 

Exam'ple. — There is a circular sharp-edged orifice in a tank 
in which there is liquid kept standing to a certain height. A 



518 AtrasD wExmASJCR. 



IB told ^ttxk, wUiMNii int^^ing nith the actual edge of 
Uie hffAe, he is allowed to do what he j^eaaes to inereaae tike 
flow. How may he do sol ETidently by fitting on m tobe 
whid& fipieklj favt giadnallj gete to be <tf mndi laiger 
diameter. He tafees eaie that the tobe shall ran foIL 

Maoample. — A nmnber of raill-ownerB leeeiTe wato* from 
the same lake. Eaeh has a reetangular cfpemMtg the depth ai 
whidi beiow the lake and its breadth are sn^iased to fix Ids 
ED^lj. Show that if a millrowner is allowed to do anytliiiig 
he pleases on his own side of tibe opeoii^ he may procure a 
very mndi greater sop^^ of water. 

419. We know that I08B df eaergy ps pound by friction, 
in water, is pr upo riA wial to the square of die wekxitx. at such 
speeds as are eommc^ in pampa. In Artt 48 we gave the 
TvHes by wfaicii we are aUe to ealmlate the loss of o^ragy 
wfaicfa a pound <tf water experiaiees in. going aloi^pipesL But 
we wiib to imprest on you tdie £giet that idiis loss always be- 
eomes ^ety gr^ when ihe flow of the wata* has to oeenr ab- 
normalfy. In the ^pe (F%. 388) ibe wide and nanow parts 
giadaaDy ^lange into one another fay 
: ■ omtimioas enr^es. It is {»aeiieally 

impoaable tor a liqmd to ^ow in a 
disecHitiniMias eorre. Siq^oee we try, 
th@i, to make it ^ow alcmg ihe pipe 
shown in fig. 290. What iha water 

^___ doesisdiis: wheniteomestotheeooier 

•nz. 2?Q. it pxidnees for itsdlf wheelil, little 

ed^es or whidpools, as we mi^t pot 

roOerB under a log of wood tiiat we wanted to get along 

eaaly; and iheste is great loss of energy doe to diiB, fin* 

the eddies have not only to be in the eomers, bat there 

are smaflo* eddies earned ahmgby 

=:::^ the water its^ maintained so long 

■'- ^""\\ ** ^^y ^'^ needed. We hare been 

J L ^eakb^ o£ actual disoontdmiity in 

Rg. 2PL ihe flow. Bat tli^e is a fact whieli 

many hydimidie engiiifiCTB seem to 

be qnite ignonat of — namd^, that a Hqaid cannot flow along a 

path wliich soddaitydbai^esin eorratare. A liqnid cannot, for 

esan^le, flow ahmg this path (Fig. 291). At a it dianges 

frcm a strai^it line sodd^ily to ihie arc of a cirde, and, oon- 

seqoaitiy, Ihe wat^ digresses at A, and aeates litde eastoi;^ 




APPLIED MECHANICS. 



519 




Fig. 292. Fig. 293. 



little eddies to carry it by a path of continuous change of 

curvature from b to c. Now this analogy can be shown to be 

true experimentally. Thus it has been found that if water 

flows along the bend (Fig. 291) it loses a certain amount of 

energy on account of the bend ; but if 

we make the pipe bend as much again in 

the same direction as in a b (Fig. 293), 

we do not get again the same loss ; 

indeed, there is comparatively very little 

loss at the second bend. But if we 

bend the pipe in the opposite direction, 

as in CD (Fig. 292), there is as much 

loss at the second bend as at the first. 

The little wheels, or castors, or 
eddies which the fluid creates for itself to 
carry it through the bend A are available 

when the water needs them again at b, where? s the wheels or 
eddies produced at c have to be destroyed, and new ones 
created, rotating just in the opposite direction to carry the 
fluid through the bend d. 

We see now how necessary it is that all curved vanes or 
other surfaces along which water flows should be drawn, not 
with a pair of compasses, but rather with a batten, a thin 
strip of wood, which bends gradually. 

420. We regard a pump as a contrivance which gives to 
every pound of water passing through it an additional store of 
energy. From the pond to the entrance to the pumj:f, every 
pound of water has just the energy it has in the pond, barring 
fvictional loss. From pump to cistern, every pound of water 
has an additional store of energy, and it is the pump which 
gives it this store. 

Suppose we have a centrifugal pump going at a regular 
speed, and discharging a regular quantity of water. ' In the 
supply-pipe a pound of water has a certain total amount of 
energy which we know, if we know its height above datum, its 
pressure, and its velocity. But, in passing through the wheel, 
it receives a supply of energy. The total energy of a pound 
of water in the discharge-pipe is greater than what it is in the 
supply-pipe. We can make all sorts of calculations, if we 
know what is the clear gain, the clear gift of energy it gets in 
passing through the pump. 

421. Suppose a man jumps into an American railway train 




520 APPLIED MECHANICS. 

anywhei^e, and after wandering about, fore and aft, jumps 
out again. Find the man's momentum in the direction of 
the train's motion just before he alights on the train. Find 
his momentum in the same direction 
when he has just sprung from the 
train , the difference of these is the 
total impulse with which he acts on 
the train. It is the momentum which 
he gives to the train. Suppose that a 
Fig. 294. number of people could perform this 

acrobatic feat every second with the 
greatest regularity, then the momentum given in one second to 
the train could be calculated. But momentum given per second 
is what we call force ; hence we have found the force acting 
on the train due to these jumping individuals, and this force, 
multiplied into the space passed through by the train in one 
second, gives the propelling work done upon the train per 
second. We have nothing to do with w^hether it is a pro- 
pelling force or a retarding force. In the one case, the 
acrobats give momentum to the train ; in the other, the train 
gives momentum to them. The loss of momentum per second 
in a regular stream of people, going on and off the train, is a 
force which is applied to the train. We only have to do with 
their momentum in the direction of the train's motion. 

Now suppose that, instead of its being a train, it were a 
sort of circular turntable, or a merry-go-round, and that a 
regular stream of people jumped on and off. In this case, the 
place where a man jumps on may be going at a different speed 
from the place where a man jumps off; but our rule is not 
very different. Find how much is added per second to the 
momentum of the wheel at the point where people leap on, 
and regard this as a force. Multiply by the speed of the 
wheel there, and this is the work done by the mere leaping 
on, and staying on the wheel. Now, find how much per 
second Ls taken from the momentum of the wheel at the place 
where leaping off occurs, and regard this as a force opposing 
the motion. Multiplied into speed at the leaping-off place, we 
have the work taken by the stream of people from the wheel, 
per second, because they leap off. The difference in these 
two things is, of course, the work done in the jumping 
on and off. 

I^ow, the water moves from the centre towards the vanes 



APPLIED MECHANICS. 521 

of the centrifugal pump, merely radially, and hence the water 
entering the vane cannot add to or diminish the momentum of 
the wheel just there. It had no momentum of its own 
previously in this direction. What occurs inside the wheel 
now we have nothing to do with, excepting that we know that 
frictional loss occurs there. We are only concerned with how 
the water leaves the wheel. The way in which it is made to 
leave the wheel determines how much energy it takes from 
the wheel. 

Take the simplest case. Suppose the vanes to be radial at 
B (Fig. 294). That is, besides moving outwards radially at b, 
the water leaves the vane with the same tangential velocity as 
the vane at b has. Suppose this tangential velocity to be v. 
Then w lb. of water leaving b per second leaves with a tan- 
gential momentum, wv -i- 32*2, and retards the wheel with a 
force of this amount acting at b. This force x -w is the energy 
which it receives per second from wheel, or wv^ -^ 32'2. One 
pound of water, therefore, receives the energy v^ -^32-2 from 
the wheel in passing through it. 

We understand, then, that a pound of water in the dis- 
charge-pipe of the centrifugal pump has this greater store of 
energy than a pound of water in the supply-pipe, except for 
frictional losses. If we make the water go out from the wheel 
as backward bent vanes make it go (Fig. 296), and as it 
would be dangerous for us to do from a railway train, less 
work has been done upon it by the wheel. If it goes out in 
the direction of motion relatively to the wheel, more work is 
done upon it than we are now supposing. 

422. The wheel with radial vanes gives v"^ -~ 32'2 foot- 
pounds of energy to one pound of water. If we know how 
many pounds of water pass through the wheel, we know then 
the total amount of work done by the wheel. 

The water gets this energy to squander or store as it 
pleases, and it does squander it in friction to a large extent. 
But suppose it squandered none of it, but converted it all into 
potential energy in lifting itself up to a cistern, it would lift 
itself v^-r- 32-2 feet high; that is, it would lift itself above 
the pond to twice the height due to the velocity of the rim 
of the wheel. Suppose the rim of the wheel has a velocity of 
45"7 feet per second, a stone would have to fall freely 34 feet 
to acquire this velocity, and hence the total rise of water 
would be 68 feet, twice the height due to the velocity of the 



522 APPLIED MECHAyiCS. 

rim of the wheeL In this case we should sav that the ptimp 
was perfect. 

The wheel itself receives energy from the enjiQe, else it 
could not give ^lergy to the water. It gives out all the 
energy that leaves the engine, except what is wasted in bear- 
ings eveiywhere, and what is wasted in friction with the 
water. 

The energy given ont by the engine per pound of water, 
divided into r^ -=- 32-2, is the efficiency of the shafting, belt- 
ing, and wheeL Agaia, the real height to which water is 
lifted by the pump, divided by the ideal height, i- -^ 32 2. is 
tiie efficiency of the water passages from pond to wheel and 
from wheel to cist^rcL 

TSie loss in these places is due to friction. Make the 
supply-pipe wide, bell-monthed at the bottom, where water 
enters it, so that it may enter by gradual curves ; make the 
approach to the wheel as gradual as possible ; let the vanes of 
the wheel make the calculable angle with the central circle, 
which will reduce the shock there (see Art. 428) ; make the 
discharge-pipe wide, and let the velocity with which the water 
enters the upper cistern h>e as small as possible, and we greatly 
reduce the waste of energy. But there is one partictilar place 
where there is usually much greater waste than anywhere 
else, and that is the chamber outside the wheel 

423. Just when the water leaves the wheel a large portion 
of its energy is kinetic. It is in rapid motioiL Xo^-, in the 
large discharge-pipe there may be as little kinetic energy as 
we please. Hence, from the time the water leaves the wheel 
till it enters the discharge-pipe there ought to be great care 
taken in allowing the kinetic energy to become converted into 
pressure enerary. 

-p Professor James Thomson discovered 

^\^ /^ here the efficiency of a whirlpool cham- 

\ / her. When we let water escape from a 

vash-basin, we know that the surface of 

^_ the water takes a shape like this (Fig. 295), 

'^j The velocity of the water is greater 

j the nearer it is to the centre. The pres- 

ife. 295t ^'"^ ^ greater the farther away from the 

centre. The spiral motion which we ob- 
serve in this case is the only steady motion of water which 
allows a constant radial discharge without the water making 



APPLIED MECHANICS. 523 

objections, setting up little eddies of its own, and thus wasting 
energy in friction. 

Hence, in Thomson's pump, after the water is discharged, 
it circulates in this cylindric whirlpool chamber, which he 
made of twice the diameter of his wheel. When a pound of 
water reaches this place, in consequence of its radial and 
circular motions, it retains more nearly the whole of its total 
energy than if we let it discharge in any other way. It has 
lost much of its velocity, but has gained in pressure. This 
whirlpool chamber of Thomson's, then, did for the water what 
gradual curves do for the water in a pipe ; it enables the 
water to convert its kinetic into pressure energy, with a mini- 
mum of waste in friction. 

It has, however, to be remembered that even here, at 
the outside of the whirlpool chamber, the water retains- a 
very considerable amount of kinetic energy, even when the 
whirlpool chamber is made very large, and much of this is 
wasted afterwards. And, although no one believes more 
firmly in the reasoning of Thomson than I do, I feel that 
perhaps the whole problem admits of a better common- 
sense solution. Let the whirlpool chamber get wider or 
broader, as well as larger in diameter. Let the wheel have 
larger orifices on its outer circumference than on its inner 
circumference. The water will lose its kinetic energy far 
more rapidly as its passage widens more rapidly. The result 
of this will be that although in this rapid change there is more 
friction, yet when the water is only a short distance out, it is 
not moving much faster than it will do in the discharge-pipe, 
and there is much less loss in entering the discharge-pipe. 
We have always thought that since Thomson's chamber is 
expensively large, we ought to sub- 
mit to a modification of his con- 
ditions even from the place where 
water leaves the wheel. But when 
we begin to consider a much smaller 
chamber, we see that possibly the 
vanes ought rather to slope back- 
wards than to be radial. Let the 
radial velocity at M n be v^. The 
velocity relatively (see Art. 36) to Fig. 296. 

the vane is v^ -^ sin. 0, if m n p is 0. 

424. Let V be the tangential velocity of the wheel at n. 





524 APPLIED MECHANICS. 

Using a very easily understood graphical method of working, let 

N represent the radial velocity at N to scale. Let m n and n p 
represent the direction of the vane and rim at n. Let n p 
represent v, the tangential velocity, and if o m is drawn at 
right angles to on, m n represents the 
velocity of the water relatively to the 
vane. Hence, as the resultant of m n 
and N p is M p, M p is the total velocity 
of the water leaving the wheel. It has 
the tangential component m q, and the 
radial component Q P which it had in the 
wheel. Suppose we call m q by the 
letter v. Now, every pound of water leaves with the tangential 
momentum v 4- 32*2, or say vjg. Every pound per second 
therefore represents a tangential retarding force, vjg, acting on 
the rim of the wheel, and this multiplied by v, or v vjg, is the 
work done usefully per pound of Avater. If we take it that in 
all cases there is a loss of the fraction s of the kinetic energy 
due to tangential motion, only s v-jg is wasted ; v vjg is total 
energy ; (v v — s v'^)lg is useful energy, or height to which the 
water is lifted. We may say roughly that the efficiency is 

V 

1 — s - ; s depends on the size of the chamber. What the 

value of t', and therefore the angle 0, ought to be, is therefore 
a question of minimum total waste of value, in interest on 
plant and waste of energy, etc. 

425. In the case of pumps we saw that each pound of water 
gets an increased store of energy, which may 
be in the shape of pressure energy, or kinetic 
energy, or both, but which mainly becomes 
potential. 

Now, in water-wheels, turbines, water- 
pressure engines, including hoists and lifts, 
we take part of the store of energy from each 
pound of water, giving it to machinery. 

As a simple case of the abstraction of 
energy from water, and as an illustration of 
the acrobat and railway-train principle, con- 
sider the vessel (Fig. 298) from which the 
water is flowing. Water leaves this vessel 
horizontally from an orifice, taking away with 
it momentum. The quantity of momentum 



f 



APPLIED MECHANICS. 525 

it takes away per second is simply the force acting on the 
vessel. You see that there is a force acting, for T have 
arranged the vessel as the bob of a pendulum. 

426. If we let the water flow from an orifice through which 
it comes in parallel streams, it is easy to show that the force 
acting on the vessel is twice the total pressure which would 
act on this little sluice when it closes the orifice, and no water 
is flowing. For if a little area a (square feet) of the orifice is 
h feet below still water level, the pressure at a being atmo- 
spheric, the velocity v = >y2gh', the volumetric flow is a v 
per second ; or the mass per second i^ w a v/g, ii w in 62-3 lbs. 

per cubic foot; the momentum per second is x v, or 

2w ah. When the orifice is closed the force due to pressure 

upon it h w a h. There have been no very sound writers on 

this subject except Thomson, but 

even the soundest imagine the 

force to be less when the vessel 

is moving. They forget in their 

calculation that the water leaving 

the vessel had at the beginning ^^^^ 

the motion of the vessel itself. ~z:. T. 



Fig. 299 shows a vessel floating on Fig. 299. 

a pond, and, moving under the 

action of its jet ; with sufficiently delicate apparatus, it may be 
shown that the force on it is the same when it moves as when 
it is at rest. If such a vessel is kept supplied with water, it is 
easy to calculate the force due to the horizontal velocity of the 
supply water ; in fact, we must consider that the acrobats enter 
the train (Art. 421) as well as leave it. Thus, in the propul- 
sion of a ship, a large centrifugal pump draws water from be- 
neath the ship, and propels it out at the sides and stern wards. 
Suppose the water moves through the nozzles with the 
velocity of 30 feet per second, and that the ship is moving the 
other way at 20 feet per second, then it is evident that the 
water has a velocity relatively to the sea of 10 feet per second. 
The momentum, therefore, given to a pound of water is 
•^V X 10, and this, multiplied by the velocity of the ship, 
gives 6J foot-pounds of energy, which each pound of pumped 
water imparts to the ship. Notice that all the kinetic energy 
is wasted, or J 3^ x 10^, or 1-56 foot-pounds of energy per 
pound of water. 



526 



APiPLiED MECHANICS. 



It is easy to see, if we had no friction in passages, that the 
greatest efficiency is arrived at by letting the water take with 
it only a very small amount of kinetic energy as it mingles 
with sea water ; that is, by letting the backward nozzle 
velocity of the water be very little greater than the forward 
velocity of the ship. But of course this is not at all a prac- 
tical solution of the problem to find the proper speed for 
maximum good result. 

427. A turbine, water-wheel, or water-power engine takes 
energy from each pound of Avater, and gives it to machinery. 
Suj^pose, for example, that we have water in a tank or dam, 
and we have a clear fall of 60 feet. Now, when a pound of 
Avater is nearly motionless at the surface of the dam, it has 
just 60 foot-pounds more energy than when it is nearly 
motionless in the tail race at the bottom. A water-power 
engine of any kind is constructed to abstract this 60 foot- 
pounds of energy with as little waste in friction as possible. 
Instead of being at the same pressure in the dam and tail 
race, v>e may have the pressure energy much greater before- 
hand, as well as the potential energy ; but in every case we 
try to take out of a pound of water the total difference of 

energy. Thus, suppose a 
pound of svater to be motion- 
less in a mill-dam 60 feet high 
above the tail race, we can- 
not take more from it than 
60 foot-pounds of energy. 
Suppose a pound of water to 
be motionless 60 feet above 
the tail race, but that it is 
also inside an accumulator, 
where the pressure is 700 lbs. 
to the square inch; we can take 
from it 60 + 2-3 X 700, or 
60 + 1,610, or 1,670 foot- 
pounds of work. 

As Ave understand the 
action of the centrifugal 
pump, we have no difficulty 
m uu:lerstanding the action of the turbine. It is because 
Ave have studied the centrifugal pump that we dwell upon 
the Thomson turbine. Water Aoavs from a pen -trough 




Fig. 300. 



APPLIED MECHANICS. 



527 




Fig. 301. 



through cast-iron pipes to a. These pipes must be bell- 
mouthed; they must open out gradually into the cistern; 
they must be as large in diameter as we can conveniently 
make them. In that case the 
velocity in the pipes will be small, 
and, therefore, the friction will 
be small. Fig. 300 shows a plan 
of the chamber, b, into which the 
water flows. The chamber is 
so large that the velocity there is 
small, and the water finds its way 
equally readily into the central 
space, whether it flows between 
the guide-blades 1 and 2, or 2 and 
3, or 3 and 4, or 4 and 1. We are 
at last allowing the water to flow 
quickly, for the guide-blade cham- 
ber is narrow. When the water is just leaving the guide-blades 
it flows rapidly ; of course it is flowing radially as well as tan- 
gentially to the rotating wheel, F, but the tangential motion 
ought to be equal to that of the wheel. 

428. Suppose we want to enter a moving railway train or 
tram-car without shock, we try to get a velocity equal to that of 

the train, in the direction of 
^.^ — ^^^^^ the train's motion, before we 

/ \ venture to enter the train ; 
/ ^ a hence the tangential velo- 

city of the water must be 
equal to that of the end of 
the radial vane of the wheel, 
if the water is to enter it 
without shock. If the vane 
is inclined like Fig. 302, a, 
the tangential velocity of 
the water ought to be less 
than that of the wheel just 
here. If the vane is in- 
clined like Fig. 302, b, the 
Fig. 302. tangential velocity of the 

water is made greater than 
that of the vane. In fact, the relative velocity of water 
and vane must be in the direction of the vane, if there is 




Q 



528 APPLIED MECHANICS. 

to be no shock. Usually the vane is shaped as we see it in 
Fig. 301, which is an enlarged section of the wheel, f ; but 
we shall suppose it to be i-adial just at the outside, for 
simpHcity of calculation. We must remember, then, that 
somehow or other we must try to get a tangential velocity 
of water equal to the velocity of vanes there. The water 
now flows through the wheel, which lets it escape at the 
centre. Here, again, we must rememl^er that the water has 
to escape with no velocity except a radial one. 

If we wanted to let a stone out of a railway carriage so 
that it would just fall to the ground vertically, so that it 
would possess no forward motion, we must shy it backwards, 
with respect to the train ; give it a velocity backwards as 
much as it has forwards already. These vanes, then, at the 
centre, let the water out backwards, just l^ecause we want the 
water to have no forward velocity when it has left the wheel. 
The water has, of course, a radial velocity everywhere, which 
simply depends on the total quantity flowing per second, 
divided by the tangential areas of these orifices. 

429. We want. now. to know how much store of energy each 
pound of water has lost in passing through the wheel, and we 
employ the rule already given. Fiad the tangential momentum 
of the water at f. If the velocity of the outside of the wheel 
is V, then v -^ d'2"2 is the forward momentum of one poun'i 
of water. This, mtdtiplied l-y i*. is the work done per pound 
of water, or — 

r- -^ o'2-2 foot-pounds. 
because it enters the wheel. Or w lb. of water per second 

means a momentum of ^7^ v per second, and this is force ; 

force multipHed by velocity v gives work done per second, and 
this divided by w gives work per pound of water. It is evident 
that if we take moment of momentum lost by the water per 
second in I'assing through the wheel, and multiply by the 
angular velocity, we get the same answer. Tlie wheel does 
no work on the water as it leaves at k. because the water 
leaves with no forward or backward momentum. Hence one 
pound of water, from the time it enters the wheel to the time 
it leaves, loses 

V' -f- 32-2 foot-pounds, 

fi^om its store of energy, and gives this store to the wheeL 



APPLIED MECHANICS. 529 

If, then, it loses no energy by friction anywhere, when it 
enters the tail race it has just this much less energy than 
when it left the pen-trough. If h is the total height of the 
fall, evidently one pound of water really gives out h foot- 
pounds of energy, so that h is twice the height due to v. We 
know that, in practice, what the water gives to the wheel is 

less than h, and -^ -r- A is called the hydraulic efficiency of 

the turbine. It is the ratio of the energy given to the wheel 
to the total energy lost by the water in falling from one level 
to the other. If, then, there is no shock to the water in 
entering or leaving the wheel, its efficiency is twice the height 
due to the velocity of the rim divided by the real total fall of 
the water. 

Of course all the energy given to the wheel is not utilised. 
There is friction between the wlieel-covers and the wheel-case, 
friction at all the bearings, etc., of the shafting and gearing 
which transmit the power of the wheel to a mill. We are 
only speaking now of the efficiency of the passages, which is, 
however, the most important matter in connection with 
turbines. 

Knowing the average amount of water passing through the 
wheel, and therefore the radial velocity at k, the angle of the 
vanes at k is determined if we know the- average speed of the 
wheel. If the speed and quantity of water were exactly pro- 
portional to one another ; that is, if the speed of the wheel 
were exactly proportional to the horse-power, the inner ends 
of the vanes once settled would remain right always. But if 
our wheel is to be regulated as a steam-engine, so that quick- 
ening speed causes less water to flow, then it is obvious that 
the inner ends of the vanes, although right for the calculated 
flow, are not properly shaped when the horse-power diminishes 
or increases. The loss of energy here is not, however, likely 
to be great in any case. 

It is different at the entrance to the wheel f. Unless the 
guide-blades are directed so as to give a tangential velocity to 
the water equal to that of the wheel, there is a considerable 
loss by friction at f. 

430. Suppose that less water flows through the turbine, the 
inclination of the guide-blades ought to alter, and this arrange- 
ment of links, which we see in the drawing, is for the purpose 



530 APPLIED MECHANICS. 

of making the guide-blades alter their inclinations to the 
wheeL Each guide-blade is pivote<:l at its extremity, K, and 
when one is shifted they are all shifted in position. Unless 
there is a great variation in the work which we reqiiii-e a 
turbine of this kind to do, it is not necessary to apply a 
governor which partially stops the water supply when the 
machinery rtms a little too quickly, although such governors 
are very necCvSsary for a great many water-wheels and turbines. 

It is to be remembered that this turbine is really a centri- 
fogal pump, through which the water is flowing negatively. 
Increased speed tends to stop the flow. If the wheel were at 
rest, the flow wotild be very much greater than it is. Hence, 
increasing the speed somewhat stops the flow, allows less 
water to pass through, and less work to be done. This action 
cannot be called a governor action, for it does not maintain a 
constant speed, but it may be called a steadying action, as it 
prevents any great change of spee<:l, even for a considerable 
alteration in the work done. 

Except at the speed for which the positions of the giiide- 
blades are fix:ed, there is extra loss in friction, and the gidde- 
blades ai-e rearranged should anv considerable chanoje be 
meditated in the power to be siven out. 

431. In arranging a turbine^ it is obvious that the great 
point to settle beforehand is this : — What ought to be the speed 
of the wheel for a given height of fall ? If there were no loss in 
friction, we could say at once, if 17 is velocity of rim of wheel, 
t7- -^ 32, the total loss of energy by one pound of water, ought 
to be equal to h • that is, the velocity of the wheel ought to 
be that due to haK the height of the total fall of the water. 
Thus, for a faU of 60 feet in height, half of this is 30 feet ; 
and if a stone fell 30 feet, it would be falling with a velocity 
of 44 feet per second. The rim of the wheel ought to have a 
velocity of 44 feet, then, per second, and it is easy to show 
that, wherever the turbine may be placed, whether it has a 
long discharge pipe, or is submerged, the water may be made 
to flow tangentially into the wheel with the same velocity as 
the wheel itself has. 

But we have usually to calcidate on the assumption that a 
certain fraction of the energy of the water is wasted in the 
supply and discharge pipes, and the discharge chamber, and 
hence the velocity of the wheel is less than that due to half 
the height of the fall 



APPLIED MECHANICS. 531 

It is usual to assume that the radial velocity of the water 
through the wheel is one-eighth of that due to the total fall. 
Dividing this into the number of cubic feet of water flowing, 
we know the total tangential area of the space between the 
vanes everywhere in the wheel, assuming that it is the same 
everywhere and it usually is. It is usual to take the innei 
radius of the wheel, e m, equal to the depth, m g, of the passages 
in the wheel, so that both these dimensions are now fixed. 
The outer radius is generally twice the inner one, and we have 
already calculated the tangential velocity of the outside, so 
the number of revolutions per minute may be calculated. The 
horse-power given out is usually taken to be less than three- 
fourths of the true horse-power of the water. Thus, by rules, 
partly due to practical experience and partly due to imperfect 
theory, we are 9.ble to fix all the dimensions of a turbine of 
the kind we have been describing. 

432. We think that, by entering thus fully into the theory 
and construction of Thomson's turbine, we can dispense with 
giving a catalogue of the constructions of turbines generally. 
This turbine is said to be one of " inward radial flow." We 
see that, for a given quantity of water flowing, it can be made 
hydraulically perfect ; that is, by proper construction of the 
guide-blades, there is no necessary loss of energy, any more 
than in the whirlpool chamber of Thomson's centrifugal pump. 

In the same manner, we could discuss the action of water 
in the unsteady " outward radial flow turbines," and, again, 
in the axial-flow turbines of Fourneyron and others. The 
principle of our stream of acrobats jumping on and ofi" a 
merry-go-round will in every case tell us how much energy 
the water gives to the wheel of a turbine, whatever may be the 
nature of the flow. In much the same way, also, we consider 
the construction of the floats of undershot water-wheels, and 
all other wheels on which the water acts impulsively. 

In the same way we might discuss a steam turbine, or the 
action of air in motion on windmills. Steam turbines are dis- 
cussed in my book on the Steam Engine. 

When the available fall is over 200 feet, it is not advisable 
to use a turbine water-wheel. In the turbine, as we saw, 
there is at least one part of the arrangement in which about 
half the total store of energy is in the shape of kinetic energy; 
and when the energy is in the shape of kinetic energy, there 
is a great waste by friction. The waste is proportional to the 



532 APPLIED MECHANICS. 

kinetic energy — that is, to the total energy — and hence tnrbinea 
are at least not more economical on very high falls than on 

moderately low ones. 

433. Steady Motion in Fluids. — The mathematical expre^cfn. of 
the law 'Art. 410 is true wliethCT^ the motion is steady or not, but we 
gire to the mathematical exptiesaom. fbs eneryy meaaoing niioi ibe 
motion is steady. Motioa of a fluid is steady when erory paztide 
at any point mores in esadOj &b same 
way as an its predeoesaara t£se. The 
SDCcesdre portions ocf a particle waA 
Out a stieam. liney and a uuiidie ox 
stream. lines He in a stream tnhcL Left 
A B he a TQty thin stream tube. Ate 
let &e pressure he j9, the Telocity in the 
directim towards b he r, and let c he * 
Zr~-. Trr! "y ahore some datnm lerd. 
I. :ir^it to the oaitre line CD 

Pig. ?:: 'J ^ rl^ 2 with the TerticaL 

It - ; - !f. A -f 5A he these 

qTiantitle; :•: r. : i Ir: i " " ;:- :i; : Irt tLe sectioa at c oar 

at D l)e " [ r :: _ Ir _ .i :i . : :i ^r -:: :hr portion of flirid 
in the r : - - - : i - - :: i. ?^: - r the tiihe. The 




if f r is 

of tie 



15 :. 
di-: 



Sp;a-^.:z. -,:;^ .= . ^. 

— : = r . S-r. 



z = 5A. we find 



: : - -^ - 5,^ = 0. 

Or, as we may write it, witl :ir : t :':.:.: It vahies oi 5r, etc. 
are smaller and smalleT wltl - ^im:, 

—j- - - - = ^ 0). 



APPLIED MECHANICS. 533 

For liquids, w is constant, and hence, integrating, 

— — i- — + h = constant .... (2). 
1g w ^ ' 

If «^ is not constant, we cannot integrate until we know how w 
Taries as a function of p. Until we know tliis, we can only write 

■ , — + A = constant (3) 

2g 1 ... \ J 

We note in (2) that v^l2g is the kinetic energy of a pound ol 
water, h is its potential energy. And, be it because of Art. 410, 



is the kinetic energy 
rgy. And, be it becat 

I- 



or merely as a help to the memory, we mean to caU 1 dpjw, or pjw, 

the pressirre energy per pound of fluid. The total energy of a 
pound of fluid remains constant, except in so far as friction 
may diminish it. 

Exatnple. — In Fig. 306 we see some stream lines of a fluid 
leaving a vessel by an orifice. We assume no friction. I shall 
consider the various orifices shown on an enlarged scale in Fig. 286. 

We wish to make the statement that (2) is the same at a as at 
c. Observe that at a the pressure is atmospheric. But what is 
the pressure at c ? If c is a stream line touching the atmosphere, 
of course the pressure is known; but inside the stream at c the 
pressure is not known with certainty. We assume it to be 
atmospheric. Now a pound of water in coming from a to c has 
had no change in its pressure energy. It has lost potential energy 
H feet, if H is the vertical depth of c below a. It had no kinetic 
energy at a ; and if v is its velocity at c, its kinetic energy there is 
^2/2 g. Therefore we say that the loss h is equal to the gain v^jl g, 
OTV — sj "2, g y.. 

It will be noticed that we choose c in Fig. 286 at places where 
it cannot be very wrong to assume atmospheric pressm^e, and 
where the stream lines are all presumably normal to the cross- 
section a. Hence the total quantity of water flowing per second is 
0, = a fJ2g'H. . . . . (4) cubic feet. Our two difficulties, what is 
a ? what loss is there by friction ? are solved by experiment. It 
is interesting to know that the most careful measurements of a and 
Q, when orifices are sharp-edged and water is flowing, show no 
perceptible loss by friction. This is one reason why sharp-edged 
orifices are preferred in the measurement of water and other fluids. 
Another reason for this is the accuracy with which we can verify 
the shape of an orifice. Hence in square, or round, or triangular- 
shaped orifices we have taken great pains to find such a place as c, 
and to find a there. In the case of a round orifice a equals the 
area of the hole multiplied by 0*62 ^^ery nearly. The part c is 
called the " contracted vein " often referred to by writers. In the 
case of Fig. 305, a is half the area a of the small hole m x. This 
is the only case in which we can make any easy attempt to 
calculate the area of the contracted vein. Note that mth this 
orifice, whatever be the shape of the vessel, the total pressure of 



534 



APPLIED MECHANICS 



the fluid upon it in the direction opposite to the arrow at c is 
known to us, being a m; h if h is the depth of the centre of a below 
water-level, because the velocities, and therefore the pressures, 
everywhere, except near mn, are the same as if the orifice were 
closed ; and near m n there are no pressure-forces parallel to the 
arrow at c. Hence Awn is the momentum leaving the vessel per 
second. It is only when the orifice is small that we can be sure 
that the average velocity at c is ^ 2ff h, and the volume flowing 
per second is a J Ign. The mass per second is this multiplied by 
wlg^ and the momentum per second is this multiplied by ^J ^gn; 
and hence awu = 2awB:, or a = 2a. 

434. For all other sharp-edged orifices than that shown in 
Fig. 305, we rely upon experiment. Q the quantity in cubic feet 
per second flowing from a sharp-edged orifice of area a square 
feet, the centre of the orifice being h feet below still water 
level, q = k y/ 2gB.. In Fig. 305 k is J, as we have seen. 
For circular orifices, k is 0*62 very nearly, even when the 
upper edge of the orifice is comparatively near the upper level, 
so long as the orifice keeps filled. Again, for square and 
rectangular orifices K is very nearly 0-62. We shall not give 
the great table of numbers, varying from 0-61 to 0-63, which 
have been experimentally determined for various sizes and 
positions of rectangular orifices, because we do not think it 
more accurate than the statement that 0*62 is very nearly 
correct for all. 

435. The shape of the stream coming from a rectangular 
orifice is very interesting ; and a student must meditate, when 
looking at such a stream, upon the way in which its component 
parts collide to cause the curious palpitating change of shape 




Fig, 305. 



Fig. 304. 

and section which is going on. Fig. 306 gives some notion of 
these changes. 

436. Experiment has shown that in the case of sharp-edged 
orifices there is no practical difference in the actual flowing of 



536 APPLIED MECHANICS. 

water from what the flow would be if the liquid were fiiction- 
less. It can be shown that when liquid is frictionless the 
stream lines from similar and similarly placed orifices in 
similar vessels with the same kind of liquid at similar heights 
are similar, the corresponding velocities being proportional to 
the square roots of the dimensions, and therefore the volumes 



A- 



Fig. 307. 



flowing being proportional to the two and a half powers of the 
dimensions. If, then, water flows from a pond over a sharp- 
edged notch shaped like a right-angled isosceles triangle, each 
of the edge5 making 4:0' with the hoiizontal, as in Fig. 287, 
and if the difference of level from b to a is h, the quantity q 
flowing in cubic feet per second is proportional to h^. Prof. 
James Thomson gave us the above principle and this method 
of measuring water. By careful measurement, he found that, 
H being in feet, Q = '2-635 Hf ■ • • (1). The rectangular 
notch is more convenient. Professor Jas. Thomson showed 
that the empirical formula of Mr. Francis, of Lowell, arrived 
at with great care and at great expense, is a rational one. 

If L is the length, of the notch in feet, h being vertical height 
in feet from sill b to still-water leveL for a given h there is a 
certain value of l beyond which increase in l means that the 
increase in q is proportional to the increase in l. In fact, we 
distinguish the flow through the two ends of length m h at 
one side and m h at the other, and the flow through the middle 
part L - 2 ms. where all the lin^ of flow may be regarded as in 
Tertical planes. We have good reason to take m to be constant. 
Imagine an orifice of length 2 m h. The flow through it is 
AtiH?, where I:i is some constant. The flow through a square 
orifice of height h, the Hues of flow being in vertical planes, is 
A-jhI, where Ag is 5«jme constant and therefore the middle flow is 
(l - 2 m h)/h tim^ this, or 

, 5 , 7 L - 2 m H 5 

H 

This will be found to reduce to q = J (l - <; h) n I. If there is 
only one end contraction, e is evidently halved : and if there are no 
end contractions, c is 0. 

The experiments of Mr. Francis give us the values of 
b and c, so that ^^ 

Q = 3-33 (L - iQH)Hl (2^, 



APPLIED MECHANICS. 537 

where rj, is 2 or 1 or 0, according as we have all the edges 
sharp or we have the edge b c a smooth vertical guiding plane, 
or both B c and d c smooth vertical guiding planes. 

437. In a gas, we have w ex p, where p is the pressure in 
pounds per square foot, if the temperature could he kept con- 
stant, or we have the rule for adiabatic flow lo <x p^ ^ where y 
is the well-known ratio of the specific heats. In either of these 

cases it is easy to find I — and write out the law. This law is of 

universal use in all cases where viscosity may he neglected, and is 
a great guide to the hydraulic engineer. Thus in the case of 

— (fp 

adiahatic flow, w = cpy , the integral of — is 

v^ 1 7 1 — — 

and hence A + 77- + r p y = constant .... (4). 

In a great many problems, changes of level are insignificant, 

and we often use v^ ■] — ^ p y z= constant . . . , (4) for 

c y — 1 ■^ 

gases. Thus, if Pq is the pressure and ivq the weight of a cubic 

foot of gas inside a vessel at places where there is no velocity, and 

if outside an orifice the pressure is^, the constant in (4) is evidently 

2 7-1 

-f — — —- v<. T , and hence outside the orifice 

c y - \^^ ' 

7-1 7-1. 



2^ 



1 (^0 ^ - i? ^ ) . . . . (5) 



and as c is Wf^ -^ p^ y, it is easy to make all sorts of calculations on 
the quantity of gas flowing per second. Observe that \i p is very 
little less than p^ and if we use the approximation (1 + «)" = 1 + na, 

when a is small, we find v^ = — [pf, - p) . . . . (6), a simple rule 

which it is well to remember in fan and windmill problems. In a 
Thomson water turbine the velocity of the rim of the wheel is the 
velocity due to haK the total available pressure ; so in an air 
tiu"bine, when there is no great difference of pressure, the velocity 
of the rim of the wheel is the velocity due to half the pressure 
difference. Thus, if ^^ of the supply is 7,000 lbs. jDer square foot, 
and \fp of the exhaust is 6,800 lbs. per square foot, and if we take 
Wq = 0-28 lb. per cubic foot, the velocity of the rim v is, since the 
difference of pressu re is 200 lbs, per square foot, 

j^l (100) = 151 feet per second. 



v; 



538 APPLIED MECHANICS. 

Returning to (6) : Neglecting friction, if there is an orifice of 

area a near -which the flow is guided so that the streams of air are 

parallel, q, the Yolume floTvdng per second, is q = v a; and if the 

pressiu-e is p, the "weight of stuff flowing per second is w = v a w 

1^ 

or V A cp'i . Using v from (5), and letting p/p^) he called o, we have 
after simplification 

W = A a^ p, V i^-^ "Vl - a '7"). 

Frob/em. — Find p, the outside pressm-e, so that for a given 
inside pressure there may he the maximum flow. 

It is oh^dous that as jl? is diminished more and more, v, the 
velocity, increases more and more, and so does q. But a large q 
does not necessarily mean a large quantity of gas. We want w to 
he large. When is w a maximimi r That is, what value of a in all 
will make 

a-fil - a t \ or a t - a 7a maximum ? 

Differentiating with regard to a, and equating to 0, we have 
2 1 



Dividing hy a t , we find a = ( —-:^ — y ' 



7 \ 7 



In the case of air 7 = 1*41, and we find p = -527 pQ. That is, 

there is a maximum quantity lea-^"ing the vessel per second when 

the outside pressure is a little greater than half the inside pressure. 

Frohlem. — When^ is indefinitely diminished what is " ? 

Answei' : v = \ / llLJL J^ 



V.^ 



7 
This is greater than the velocity of sound in the ratio 

being 2'21 for air. That is, the limitino- velocitv in 



^ _ 1 ___ 

the case of air is 2,413 feet per second x A / __, where t is the 

V 273 
absolute temperature inside the vessel, and there is a vacuum outside. 
This involves the idea of the jet creating such intense cold as to be 
at the absolute zero of temperature. 

Returning to equations (3) and (4), we assumed h to be of little 
importance in many gaseous problems of the mechanical engineer. 
But there are many physical problems in which it is necessary to 
take account of changes in level. For example, if (3) is integTated 
on the assumption of constant temperature, and we assume v to 



APPLIED MECHANICS. 539 

keep constant, we find that 2^ diminishes as h increases, according 
to the compound interest law. Again, under the same condition 
as to V, but with the adiahatic law for tv, we find that y diminishes 
with h according to a law which may he changed into " the rate 
of diminution of temperature with h is constant. " 

A great numher of interesting examjDles of the use of (2) might 
be given. It enables us to understand the flow of fluid from 
orifices, the action of jet-pumps, the attraction of light bodies 
caused by vibrating tuning-forks, why some vah^es are actually 
sucked up more against their seats instead of being forced away b)'- 
the issuing stream of fluid, and many other phenomena which are 
thought to be very curious. 

438. Example. — Particles of -water in a basin, flowing very 
slowly towards a hole in the centre, move in nearly circular paths, so 
that the velocity v is inversely proportional to the distance from the 

centre. Take^v = -, where a is some constant and x is the radius 

X 

or distance from the axis. Then (3) becomes h + - — ;, + — = c 

2gx^ w 

Now at the surface of the water p is constant, being the pressure 

of the atmosphere ; so that there h =■ - - — ^, and this gives us 

the shape of the curved surface. Assume c and a any values, and 
it is easy to calculate h for any value of x and so plot the curve. 
This curve rotated about the axis gives the shape of the surface, 
which is a surface of revolution. 

A student who depends upon a text-book to give him complete 
information is not learning to become an engineer. Meditation 
when looking at water flowing from a basin ought to gTcatly add 
to what he will obtain from such a book as this. Perhaps he will 
begin to notice that it is centrifugal force due to whirling motion 
which maintains pressure at a place. What will be the effect of 
friction at the solid surface of the basin ? It will diminish velocity 
and diminish pressure. Water will flow, therefore, down the 
surface of the basin and towards the hole. If this is well under- 
stood, the student will understand how it is that at a bend of a 
river the earth from the outer bank is dragged along the bottom 
and deposited on the inner bank, and hence that a river through 
an alluvial plain is always tending to get more crooked, until at 
length it cuts off a bend and so straightens itself in a new channel. 
In the basiu problem we have also James Thomson's explanation 
of the phenomena of great forest fires, and also of the prevailing 
wind system of the earth. (See Phil. Trans.) 

439. If water flowing spirally in a horizontal plane follows the 

law V = -, where r is distance from a central point; note that 
r 

W IP' 

p = c - \ — -^. The ingenious student ought to study how p 

and V vary at right angles to stream lines. He has only to 
consider the equilibrium of an elementary portion of fluid, c u 



540 APPLIED ^[ECHANICS. 

(Fig. 303), subjected to pressures, centrifugal force, and its own ' 
weight in a direction normal to the stream. He will find that if 

dp 

-f- means the rate at which p varies in the direction of the radius 
dn 

of curvature away from the centre of curvature, and if o ia the 

angle u c e (Fig. 303), the stream heing in the plane of the paper, 

which is vertical, and if r is the radius of curvature, 

dp to v^ 

-— = w sin. a . . . . 11). 

dn g r ^ ' 

If the stream lines are all in horizontal planes, 

dn = '^r-"-}^^ . 
Stream lines all circular and in horizontal planes in a liquid, so 

that h is constant. Jl v = -, where J is a constant, 

dr ~ g ' 1^ 

p = - i — 2 + constant .... (3). 

We see, therefore, that the fall of pressure as we go outward is 
exactly the same as in the exercise at the end of last article. 

Example. — Liquid rotates about an axis as if it were a rigid 

body, so that v = br, then -^ = — IPr, p = ■-- Vh^ + c. This 
dr g ' ^ 2g 

approximately shows the law of increase of pressure in the wheel 

of a centrifugal pump when full, but when delivering no water. 

It is the answer already obtained (Art. 175). 

EXERCISES. 

1. The pressure at the inside of the wheel of a centrifugal pump is 
2,116 lbs. per square foot; the inside radius is 5 foot; the oiitside radius 
1 foot. The angular velocity of the wheel is Zi = 30 radians per second. 
Dmw a curve showing the law of p and /- from inside to outside when 
very little water is being delivered. If the water leaves the wheel by a 
spiral path, the velocity everywhere outside being inversely proportional 
to /•, draw also the curve showing the law of p in the whirlpool chamber 
outside. 

2. The expression ^.o -j^ 

^r-+-p + h = 'R . . . . (4), 

which remains constant all along a stream line, may be called the total 
store of energy of 1 lb. of water in the stream if the motion- is steady. 

f/E \ dv \ dp dh ^ , J- /ix 

Now —- — -v-r--K — T-+^- becomes from equation (1), 
dn g dn w dn dn 

dB 2v 1 /v dv\ ,,, 

n being in a normal direction away from the centre of curvature and t 
the radius of curvature. This expression o ( ~ "*" rf~ ) ^^ called the 



APPLIED MECHANICS. 541 

"average angular velocity" or " the rotation'''' of the liquid. Heace 

-^ = — X rotation .... (6). 
cln g 

3. Show that the law (1) for a gas under adiabatic conditions, heing 
4 of Art. 437, the above laws (5) and (6) hold for a gas as well as a liquid. 

4. Circular Stream Lines. — What is v as a function of r, the radius, 
if the flow is to be irrotational — that is, if a pound of water or of gas has 
the same energy in one stream line as in another ? Here vjr + dvfdr = 0, 
or drjr + dvjv = 0, or log. r + log. v = c, b, constant, ox v cc. l/r. 

5. In a gas flowing irrotationally in circular streams, iS. v = -, find^^ 
everywhere. Inserting v = 5 in (4) of Art. 437, we find 

' f y-l Pw, /I 1\\t/7(-1). 

6. A centrifugal fan makes 3,000 revolutions per minute. It is 2 feet 
in inside diameter, 4 feet outside, and there is a whiiipool chamber outside 
the fan. No air is being delivered from the chamber. The pressure just 
inside the wheel is 1 atmosphere, 2,116 lbs. per square foot, and 0" C. 
Draw curves showing the pressure at any distance from the centre of the 
wheel, both in the wheel and the whirlpool chamber. The air follows the 
adiabatic law. If the speed is 40 revolutions per minute, and the fluid is 
water, draw the curves. 

440. When liquid flows by gravity from a small orifice in a 
large vessel, where at a distance inside the orifice the liquid may be 
supposed at rest, it is ob^dous the e is the same in all stream lines, 

so that -i- is 0, and there is no "rotation" anywhere. It can be 
d7- ' -^ 

proved that if a portion of frictionless fluid possesses "rotation," 
this can never be destroyed. Nor can rotation be created. But 
the student must refer to books on hydrodynamics for further 
information. When imstable states of motion set in, the mathe- 
matics get beyond the author's powers, even in straight pipes. 
The papers of Professor Reynolds may be consulted. As for what 
occurs in ^dscous fluid at a bend in a pipe, nobody has done more 
than guess as yet. At small velocities, we have a fairly good 
knowledge of what happens. Suppose very viscous liquid is 
flowing along a straight pipe, the velocity everywhere is such 
that the loss of energy by viscosity is a minimum. If the pipe 
has a curved centre line, the distribution of velocity everywhere 
is different from what it is in the centre line. 

When fluid flows in circular cylindric stream lines, a stream of 
internal and external radii r and r + 5r, of imit breadth, is acted 

on by tangential force on its internal cylindric sm^face H- {-j , — ; ) 

per unit area, if v is the velocity. These forces act on the area 
2 Trr at a distance r from the axis, so that their total moment' 

'dv 



„ = 2.„.^(|-?)....(l). 



542 APPLIED MECHANICS. 

Let the moment of the forces on the outside surface be m + 5 m. 
If there is no alteration of pressure along a stream line, and only 
balanced forces in planes at right angles to the axis, then 5 m is the 
only cause of increase of moment of momentum of the stream. 
The state into ^'hich the system settles is one in which the motion 

is steady. In fact, ^ is 0, or 

dPv I dv V . ... 

3-0 + - -7 2 = • . . . (2). 

dr^ }• dr r^ ^ ' 

The general solution of this is 

t; = Ar + ^ (3). 

If at a cylindric siuface r = Vq, v = Vq, and i£ at another r = rj, 
v = 0, then 



.(4). 



So that B = - A r{\ t'o = A (ro - ^), a = J "^" „ , and hence (3) 

becomes v = J"'" ^ (''- — )■ If, then, viscous fluid escapes 

fromthe^ rim of the wheel of a centiifugal pump with little radial 
velocity into a chamber boimded by parallel frictionless faces, if r- 
is the outer radius of the chamber, 







(5). 



And if ri = 00, then a must be in (3), and Cq = --, so that (3) 
becomes 

« = '-f2....(6). 

The loss of energy per second is - [iv I J^ - - ) per imit Tolimie, 
or - yu ' ^^';'' ( - 2 ' ^',^") = 2;u ^'^l^~. The volimie of the ring between 

•i TTyu? o-ro'- . , is the total loss, or 

4 ttixVqVq'. If V is the small radial ^-elocity of fluid leaving a wheel, 
the weight of fluid per second lea-^ing unit breadth of the wheel is 
2 ttTq \u\ The kinetic energy of it due to its tangential velocity 

only is 2'irrQ — Vq"^. Its kinetic energy per poimd is J^ . 

Total kinetic energy ^^^ rm. i • ^- j ^ ^i j- 1 

r-j ^^ = K — • The kinetic energy due to the radial 

wasted tsnergy 2fffji, °'' 



APPLIED MECHANICS. 543 

(r 2 \ 
~ - rj 

-72" - 1 - 2 + = + j"^"^' ( ~:i~ )• "^^^^ ^^ ■^^'''^ °^ 
energy per second jDer imit volume, or 2 r-^in'^fx i~ V Total 

loss = r'4 Trr^^V ('^' - 1) ^r ( - ^' - r) 

MISCELLANEOUS EXERCISES. 

1. In a force pump used for feeding a boiler the ram has a diameter 
of 2 inches and a stroke of 24 inches. How many gallons of water 
(neglecting leakages) would be forced into the boiler for each 1,000 
double strokes (one forward and one backward) of the pump ? 

Ans., 272 gallons. 

2. How many gallons of water will be delivered per hour by a single- 
acting pump (diameter of plunger, 4 inches; length of stroke, 12 inches) 
making 24 strokes per minute, the slip being 15 per cent. ? How long 
would it take this pump to fill a tank 10 feet by 6 feet to a depth of 3 
feet? Ans., 665 gallons ; 1 hour, 41 mins. 

3. Certain machinery worked from an accumulator requires 20-horse 
power for a quarter of an hour every hour. The pressure in the 
accumulator is 700 lbs. per square inch. If 25 per cent, be allowed for 
frictional losses, find the size of a single-acting pump which, driven for 
thirty -five minutes every hour by a donkey engine, making 50 strokes a 
minute, will just have the ram at the top of its stroke at the beginning 
of each hour. Assume a slip in the pump of 25 per cent. 

Ans., '094 cubic foot. 

4. In a double-acting force-pump the diameter of piston is 12 inches 
and the stroke 2 feet 6 inches. The distance from the pump to the well 
is 15 feet, and from the pump to the place of delivery is 35 feet. Find 
the horse-power required to work the pump if 30 per cent, is wasted in 
friction and the number of strokes be 40 per minute. Ans., 21 16. 

5. The diameter of a pump bucket being 6 inches, and the vertical 
lift from the well to the point of delivery being 40 feet, find the load on 
the bucket. What horse-power will be necessary if the stroke be 15 
inches and there are 20 double strokes per minute? Allow 30 per cent, 
for all losses. Ans., 489 lbs. ; -53. 

6. The discharge from a pipe is 12 gallons a second. At a point 110 
feet above datum level the diameter is 5 inches, and the pressure 2,050 
lbs. per square foot. Find the total energy of each pound of water. If 
at a point where the pipe is 2 J inches diameter and 10 feet above datum 
the pressure is 1,000 lbs. per square foot, find the loss of energy between 
the two points mentioned. Ans., 146 ft. lbs. ; 704 ft. lbs. 

7. The volume of water passing along a pipe running full is 10 cubic 
toet per second. At one section the area is 2 square feet; at another 



544 APPLIED MECHANICS. 

place, 12 feet below the level of tlie former, the area of the crosa* 
section is Ij square feet. Find the difference of pressure at the two 
sections, friction being neglected. Ans., 710 lbs. per sq. ft. 

8. At a certain point, 15 feet above datum, in a stream flowing from 
a reservoir, the still surface of which is 1.50 feet above datum, the velocity- 
is 20 feet per second. "V^Tiat is the pressure at this point, assuming no 
loss by fiiction? Aus., 70-4 lbs. per sq. in. 

9. Find the power of a waterfall where 2,000 cubic feet of water pass 
per minute, the height of the fall being 30 feet. 

In a waterfall, 20 tons of water fall from a height of 36 feet in each 
minute, and are employed to turn a turbine which transforms six-tenths 
of the energv of the water into useful work. Find the horse-power of 
the turbine. ' A}is., 113-3 h.p. ; 2932. 

10. The vertical distance from still- wat<3r level to the Hp of a 
rectangular notch was observed to be 3 feet during an interval of six 
hours, -Go feet during the next twelve hours, and ■■1 feet during the next 
six hours, the width of the notch being 2J feet. Find the number of 
gallons passing through during the twenty-four hour?. 

Afis., 1,567,000 gallons. 

11. llie flow of water in a certain stream is measured by employing a 
Thomson's V-shaped weir gauge. The vertical distance from still- water 
level to the lowest point of the notch is observed to be lA feet. The 
water which passes through has a fall of 20 feet, and is employed to drive 
.1 water-wheel having an efficiency of 60 per cent. Find the horse-power 
which may be obtained from the wheel. Ans., 8 "33. 

12. A watei-fall is to be utilised for electric lighting. The engineer 
sent to inspect the place finds out the following data • — The water at one 
place flows in a straight rectangular channel 4-J feet wide, 2^ feet deep, 
with an average velocity of 3 feet per second. The available fall is 20 
feet, and the water-wheels to be used have an efficiency of 62 -per cent., 
the dynamo efficiency being SO per cent. Neglecting all other losses of 
energy, find approximately how many 60-watt incandescent lamps may 
be supplied. Ahs., 471. 

13. The rim of the wheel of a centrifugal pump goes at 30 feet per 
second. Water flows radially at 5 feet per second. The vanes are 
inclined backwards at an angle of 35" to the rim, "What is the 
absolute velocity of the water ? TMiat is the component of this parallel 
to the rim? If 120 cubic feet of water leave the rim every minute, find 
the tangential retarding force at the rim. What is the work done 
usefuUy per pound of water ? 

Afis., 23*4 ft. per sec. at 12'-3 with direction of rim; 22-86 feet per 
sec; 88-45 lbs.; 21-3 ft. lbs. 

14. Suppose water to flow in a steady stream with a constant total 
head of 100 feet, reckoned from the datum plane and from zero pressure. 
Determine the discharge into the atmosphere in gallons per minute from 
a pijie 2 inches diameter at a point 5 feet above the datum. A?(S., 638. 

15. An orifice 1 square inch area is made in the side of a large tank, 
at a depth of 4 feet below the surface of the water, and the issuing jet is 
horizontal If the jet falls vertically through If feet in a horizontal 



APPLIED MECHANICS 545 

motion of 5 feet, and the discharge be 1 6 gallons per minute, find the 
coefficients of velocity, contraction, and discharge. 

Ans., -97; -64; -62. 

16. Find the discharge in gallons per minute from an orifice 2 inches 
in diameter in the side of a tank under a constant head of 6 feet, 
measured from the centre of the orifice. The coefficient of discharge 
may be taken at -6. Atts., 96. 

17. In an inward flow wheel the velocity ox flow is 8 feet per second, 
the internal diameter 9 inches, and the revolutions 10 per second. Find 
the angle of the vanes at exit, so that the water may leave the wheel 
radially. Ans., 18°-75. 

18. Determine the velocity with which the water enters an inward 
flow turbine under a head of 36 feet, the speed of the periphery of the 
wheel being 32 feet ptr second. The vanes of the wheel are radial at 
entrance, the velocity of flow is constant, and the water leaves the wheel 
with no tangential velocity. Ans., 32-56 ft. per sec. 

19. What horse-power is required to drive a radial- vaned pump of 15 
feet diameter at 50 revolutions per minute when delivering- 15,000 gallon-s 
of water per minute ? What is the efficiency if the lift is 22 feet ? 

Ans., 219; -46. 

20. A stream of water, the volume of flow of which is 3,000 gallons 
per minute, has a velocity of 20 feet per second. It impinges on a 
succession of curved vanes moving with a velocity of 8 feet per second in 
a direction inclined at 45° to the direction of the stream. Determine the 
direction of the tangent to the vane at entrance, so that the water may 
impinge without shock. If the vanes are circular arcs of 90°, find the 
resultant pressure on the vanes, and the component force in the direction 
of motion. Ans., 21°'5 with direction of jet. 

21. Find the horse-power developed in a Thomson turbine which is 
supplied with 15 tons of water per minute, with a forward tangential 
velocity of 40 feet per second, equal to the speed of the periphery of the 
wheel, the diameter of which is 2 feet. The water leaves the turbine at 
a radius of 6 inches, with a backward tangential velocity of 10 feet per 
second. -L/ts., 57'27. 

22. The diameters of the inner and outer circumferences of an inward 
flow turbine are 2 feet and 4 feet respectively. The direction of the 
vanes at their outer ends is radial. Determine the angle at which the 
inner ends are arranged, supposing that velocity of flow through the 
turbine is one-eighth the velocity due to the total head, and that of the 
outer ends is that due to half the head. Ans., 19° -6 with rim. 

23. A turbine with radial vanes receives 50 gallons per minute with 
an effective head of 28 feet. Find what should be the total area of the 
inlet passages, and the velocity of the lips of the vanes for maximum 
efficiency. Ans., 1"51 sq. ft. ; 28 ft. per sec. 

24. The wheel of a centrifugal pump is -6 feet in diameter; the 
turning moment on the spindle is 12 pound-feet. If 160 gallons of 
water are raised per minute, find the mean velocity with which the 
water leaves the wheel, assuming that on entering it has no -s-elocity of 
whirl. Ans., 24*1 ft. per sec. 



646 



CHAPTER XXY. 



PERIODIC MOTION. 

441. When, after a certain interval of time, a body is found 
to have returned to an old position, and to be there moving in 
exactly the same way as it did before, the motion is said to be 
periodic, and the interval of time that has elapsed is said to 
be the periodic time of the motio?i. Thus, if a body moves 
uniformly round in a circle, the time which it takes to make 
one complete revolution is called its periodic time. 

442. When a body moves uniformly in a circle, as, for 
instance, the bob of a conical 2yendulum, if we look at it 
from a point in the plane of its circle, it seems merely to 
swing backwards and forwards in a straight line. Thus, 
it is known that Jupiter's satellites go round the planet 
in paths which are nearly circular, but a person on our 
earth sees them move backwards and forwards almost in 
straight lines. Now, if we were a very great distance away 
from the bob of a conical pendulum in the plane of its motion, 
we should imagine it to be moving in a straight line, and the 

motion which it would 
appear to have — slow 
at the ends of its 
path, quick in the 
middle — would be a 
simple harmonic mo- 
tion. To get an exact 
idea of the nature of 
this motion — in fact, 
to define what I mean 
by simple harmonic 
motion — draw a circle, 
\ o' L o" (Fig. 308), 
and divide its circum- 
ference into any even 
number of equal parts. 
Draw the perpendicu- 
Fig. 308. iars b' B, c'c, etc., to 




APPLIED MECHANICS. 



547 



any diameter. Now, if we suppose a body to go back- 
wards and forwards along AOL, and if it takes just the 
same time to go from a to B as from b to c, or from 
any point to the next, then its motion is said to be a simple 
harmonic motion. This sort of motion is nearly what we 
observe in Jupiter's satellites ; it is almost exactly the motion 
of the bob of any long pendulum or the cross-head of a steam- 
engine ; it is the motion of a point in a tuning-fork, or a 
stretched fiddle-string when it is plucked aside and set free ; 
of the weight hung from a spring balance when it is vibrating ; 
of the up and down motion of a cork floating on the waves in 
water; and of the free end of a rod of metal when the other 
end is fixed in a vice and the rod is set in vibration ; it tells 
us in all these cases the nature of the motion, when such 
motion is of its implest kind. Thus, for example, a cork 
floating on water may really have a very complicated motion, 
but if the wave in the water is of its simplest kind, the cork 
goes up and down with a simple harmonic motion. If you 
study the figure which you have drawn, and then watch the 
vibration of a very long pendulum, you will learn about this 
kind of motion what cannot be learnt by reading. 

443. Now let me suppose that the body takes one second to 
go from A to B, or from b to c, or from any point to the 
next in Fig. 308. Then the length of ab in inches represents 
the average velocity between the points A and b, and in the 
same way we get the average velocity anywhere else. Thus, 



in the figure from w 


hich the woodcut is 


drawn I find 






Velocity from 


A 

to 

B 


B 

to 
c 


c 

to 

D 


D 

to 

E 


E 

to 

F 


F 

to 




o 
to 

G 


G 

to 

H 


H 

to 

I 


I 

to 
J 


to 

K 


K 

to 

L 


8 in inclies per 1 
second J 


0-34 


l-OO 


1-59 


2-07 


2-41 


2-59 


2-59 


2-41 


2-07 


1-59 


1-00 0-34 



We observe that the velocity increases as the body ap- 
proaches the middle of the path, and diminishes again as it 
goes away from the middle. Now the increase in the velocity 
of a body every second is called its acceleration, and so we can 
observe what is the acceleration at every place. You see 
that the velocity changes from '34 to 1*00 near b in one 
second — that is, the acceleration near b is -66 inch per second 



548 



APPLIED MECHANICS. 



per second. Similarly subtracting 1-00 from 1-59 we find the 
acceleration at c to be 0*59, and so on. Make a table of these 
values, and place opposite them the distances of the points b, c, 
etc., from the centre. In this way we find from the figure the 
followinii" Table of Values : — 



Distance from c to 


Acceleration at 


Displacement divided by 
Acceleration. 


Bis 9-66 


B is 0-66 


14-6 


c is 8-66 


cis 0-59 


14-7 


u is 7-07 


D is 0-48 


14-7 


E is 5-00 


E is 0-34 


14-4 


F is 2-59 


Fis 0-18 


14-4 


o is o 


is 




G is 2-59 


G is 0-18 


14'4 


H is 5-00 


H is 0-34 


14-4 


I is 7-07 


I is 0-48 


14-7 


J is 8-66 


J is 0-59 


14-7 


K is 9-66 


K is 0-66 


14-6 



From this it is evident that when the distance of a point 
from the centre is divided by the acceleration at the point, we 
get about 14-6 in every case — that is, if we worked more 
exactly we should have the exact law that the acceleration at 
a place is proportional to the distance from the centre. This 
curious property is characteristic of the kind of motion which 
we are describing. If, again, we draw a number of figures, 
such as Fig. 308, and divide the circles into very different 
numbers of equal parts, we shall find that in every case the 
following law is true : — The periodic time of a simple har- 
monic motion — that is, the time which elapses from the 
moment when the body is in a certain condition until it 
gets into exactly the same condition again — is equal to 6-2832 
multiplied by the square root of the ratio of displacement to 
acceleration given in the third column of the above Table. 
Thus, in the Table we find the mean value of the ratio (adding 
all the quotients and dividing by their number we get 14-56) 
to be, let us say, 14*6. Now the square root of 14"6 is 3-82, 
and this multiplied by 6*2832 is 24 seconds, which we see by 
inspection is the periodic time in Fig. 308. 

444, We see, then, that if the force acting on a body and 
causing it to move is always proportional to the distance of 




APPLIED MECHANICS. 549 

the body from a certain point, and acts towards that point, 
the body gets a simple harmonic motion, and we have a rule 
for finding the periodic time. 

The acceleration is always towards the middle point — that is, 
whilst a body is leaving the middle its velocity is being- lessened ; 
when it is approaching the middle its velocity is being increased. 
The velocity at the middle is equal to the uniform A^elocity in the 
circle from which we imagine the harmonic motion to be derived — 
that is, the velocity in the middle is equal to 3-1416 times the 
distance a l divided by the periodic time. 

Suppose the body to be at q, Fig. 309, moving with a harmonic 
motion in the path a o l. Describe the 
circle, draw qp perpendicular to al, then 
p is the position of a body which has 
corresponding uniform circulfir motion. 

Let the time t seconds have elapsed 
since the point a was at o. The corre- 
sponding point p was then at c. Let the 
uniform speed of p be «; feet per second ; 
then c p = vt. Let o p = r ; let the angle 
c o p = vtlr, so that the radius o p moves 
with the angular velocity «^/r. The time 
T of revolution of p or of complete oscilla- 
tion of Q is 2 TT r -i- V, or 2 TT 4- the angular velocity, which we 
shall call j). Then o Q = o p . sin. o p q = o p . sin. cop, or, if we 
call o Q = a;, re = r sin. i)t. By Art, 19, 

velocity of q = , — rp cos. i)t, 

cP'X 
acceleration of a = -^ = - rj)^ sin. pt. 

Ifence the displacement x, divided by the value of the acceleration, 

if we neglect the - sign, is Ijp^ or —^ (since t = 2 irip). 

Notice that the acceleration - rp"^ sin. jjt is the resolved part, in the 
direction x of the acceleration of p. The speed of p is v, a constant. If, 
• then, it has an acceleration, it cannot be, nor any part of it, in the direc- 
tion of its iDath ; it is, then, in the direction of the radius. Let the acce- 
leration of p in the direction o p be a, then the resolved part in the 
direction o l is a cos. p o l = acceleration of q, or a ^m.pt — - rp^ sin. 
pt^ or a= -rp^. The acceleration of p is, then, centripetal from p 

V 

towards o, and this amount is rp^. Thus j? = -, so that o = - v^/r, 

the ordinary formula for centrifugal acceleration. 

Here we have obtained a knowledge of the centripetal accelera- 

tion — mathematicall}^ from knowing what is meant by linear 

acceleration. Conversely, suppose we know that the centripetal 
acceleration of p in the direction p o is v^jo p, then the acceleration 
of a towards the centre is this multiplied by cos, p o l ; that is, by 



550 



APPLIED MECHANICS. 



oa 



o q/o p, and it is therefore equal to — , — or o q . t^/o p^, ao 

o p o p ' 

tliat it is proportional to o Q. Also, o a diTided by the acceleration 
(towai-'ls the centre) is o p^/r- or 4 r,-!' (since 2 t". o p/i' = x). We 
are therefore led in manv wa\8 tu the rule 



Periodic time 
(5^ aho Art. 19.) 



.-, /Displacement 

'V Acceleration 



445. Example. — In Fig. 310, a is a ball of lead weighing 20 
lbs. carried by means of a spiral spring whose 
own weight may be neglected, let us suppose.* 
Find by experiment how much the spring 
lengthens when we add 1 lb. to the weight of 
A, or shortens when we subtract 1 lb. from 
the weight of a. Let it lengthen or shorten 
01 foot. E\-idently, if ever a is 01 foot 
upwards or downwards from its position of 
rest, it is being acted upon by a force of 1 lb. 
tending to bring it to its position of rest. We 
know also that if a is 0-02 foot or 003 foot 
above or below it^ r>lace of rest, there is a 
force of 2 or 3 lbs. trying to bring it back. 
We see, then, that the up and down motion of 
A must be simple liarinonic. When the dis- 
placement is, say 0*02 foot, the force acting 
on A is 2 lbs., and the acceleration of a is 
force 2 -4- mass of a ; and as the mass of A is 
20 -h 32-2, or 0-621, the acceleration of a 
is 3-22 feet per second per second when it is 
displaced 0-02 foot from its middle position. 

Xow, employing the rule given above, divide 
0*02 by 3-22 and extract the square root, then multiply by 
6-2832, and we get 0-495 second, or about half a second as the 
periodic time of the swinging ball. "When we make experi- 
ments wc find that, unless the coils of the spring are flat, and 
the rigid support of a exactly in the axis, the ball has a 
tendency to turn and vibrate laterally, which disturbs observa- 
tions if we make careful measurements of the length of swing. 

446. Example. — The Simple Pendulum. — A simple pen- 
dulum consists in an exceedingly suiall but heavy body 
suspended by means of a long inextensible thread, whose 

* "^'e really assiime that one-third of the mass of the spring is added to A. 



A 

Fis. 310 



APPLIED Mi£CHANICS. 



551 



weight may be neglected, capable of swinging backwards and 
forwards in short arcs. If the arcs are not too long, the time 
of one swing is always the same. Thus, in Fig. 311, s is the 
point of suspension, s P a silk thread, p a small ball of lead. 
p will move backward and forward along the path AOL with a 
motion which is simple harmonic, provided the thread is so long 
and A L so short that the force acting on the ball at any time in 
the direction of its motion is j^roportional to the distance of 
the ball from o. To show that this is so, resolve the verti- 
cally acting weight of the ball in the direction of its motion 
along A 0. We iind that it is not 
quite proportional to ao unless AG is 
very short, but if this slight discrep- 
ancy is neglected the force urging the 
ball towards o is the weight of the ball 
multiplied by o a and divided by s A, 
the distance from the point of support 
to the centre of gravity of the ball. 
As a matter of fact, the nature of 
the vibration does not depend on the 
weight of the ball ; but, to fix our 
ideas, let us suppose that the weight 
is 2 lbs., then the mass of the ball is 
2 -4- 32-2, and acceleration along a o is 



the force -^ mass, or 



2 X 



the force, 




divided by :^^ the mass, so that the 
acceleration is 32-2 x ao -=-sa. 

Now, our rule is to divide a o by Fig. 3ii. 

the acceleration at A, and this gives 

^; extract the square root, and multiply by 6-2832 for the 
periodic time of oscillation of the pendulum. The general 
rule for a simple pendulum swinging in sho rt arcs is then : 

Time of a complete oscillation = 6-2832 \/ ^athofpe^ndvMm 

The mass or inertia is 2 -4- 32*2 at all places, but the weight 
may be more or less than 2 lbs. at different parts of the earth 
in the proportion of g to 32 '2, and hence we find 

T = 27r y l^g 
as our general rule. 

The time of one swing is half this. The number 32 '2 ex- 
presses the effect of the force of gravity at London. At any 



552 APPLIED MECHANICS. 

other place on the earth's surface it would be different — that 
is, at different places on the earth a given pendulum has 
different times of oscillation. For instance, a pendulum 
taking 2 seconds for a complete oscillation at Paris — that is, 
taking 1 second for one swing, called a seconds pendulum — if 
swung at fSpitzbergen would gain 94 seconds per day ; and if 
swung in New York would lose 30 seconds per day, provided 
the pendulum did not alter in length in being taken from one 
place to the other. Evidently when a pendulum gets longer 
it oscillates more slowly ; hence in summer, vv^hen the pen- 
dulum of a common house-clock expands with heat, it goes 
more slowly, and in winter it goes more quickly, unless the 
position of the bob is adjusted. A pendulum which is self- 
adjusting — that is, which is so constructed that it remains of 
the same length whatever be the temperature — is called a 
coinjjensation j^endulum. 

447. Example. — In Fig. 312, b represents a strip of steel 
fixed firmly in a vice at c, with a heavy ball A fastened at its 
free extremity. Find the force in pounds which will increase 





Fig. 312. 

the deflection of A by 0-01 foot ; say that it is 1 lb. We 
know that a force of 2 or 3 lbs. will cause an increased de- 
flection of twice or three times this amount ; and as the force 
acting on the ball in any position is proportional to its distance 
from its position of rest, the ball will swing with a simple har- 
monic motion. If we can neglect the weight of the strip of 
steel, and if the ball is small in comparison with the length of 
the strip, its time of vibration may be calculated in exactly 
the same way as that of the ball in Fig. 311. Experiments 
with this enable Young's modulus e to be determined. 

448. Example. — Suppose that b c (Fig. 313) is a bent glass 
tube of uniform section containing a liquid which can mov^e 
without friction in the tube. If the liquid be disturbed 
so that the level is higher in b than in c, it will continue to 



APPLIED MECHANICS. 553 

swing about its position of equilibrium — that is, the position 
in which the hquid is at the same level in both limbs of the 
tube. Thus, if c is '01 foot below the proper level, and b is 
•01 foot above this level, the force which tends to cause the 
liquid to return to its proper level is twice the weight of the 
liquid o B. Suppose the weight of the liquid is 10 lbs. per 
foot length of the tube, then the force acting 
on the liquid is -02 x 10, or '2 lbs. If the 
whole length of tube filled with liquid is 6 
feet, then the weight of liquid which has to 
be set in motion is 60 lbs., and its mass is 
60 -=- 32-2, or 1*863 ; hence the acceleration 
is '2 -^ 1-863, or 0'107 foot per second per 
second. The displacement is "01, and, work- ^^\^ /^ 
ing by our old rule, displacement divided by ^^^^^^ 

acceleration is '0935. The square root of Fig. sis. 

this is '3058, and multiplying by 6-2832 
we get 1-92 second as the periodic time of the oscillation. 
We find it easy to prove that the liquid swings in the same 
time as a simple pendulum whose length is half the total 
length of the liquid in the tube, and that it is the same what- 
ever be the density of the liquid — that is, whether it is 
mercury or water. 

If w lbs. is the weight of liquid per foot in leBgth. of the tube, 
if a; is the displacement o b or o c, the force causing motion is 
2 xiv. (This force is multixDlied by ^ -^ 32*2 if the observation is not 
made in London.) If a is the total length of liquid in the tube, 
the weight of liquid moved is mv, and its mass is aw -^ 32-2. 

Hence the acceleration is 2 xw -. or — -' and the displacement 

9 a 

divided by acceleration is a; -r- — ^ or ^^ ; so that the periodic 
a 2^' ^ 

time is 2 ' 



2 • g' 

449. It will be observed that in all these cases of vibration 
of bodies there is a continual conversion going on of one 
kind of energy into another. At each end of a swing the 
body has no motion ; all the energy is therefore potential, 
whether it is the potential energy of a lifted weight or the 
potential energy of strained material. In the middle of the 
swing the body is going at its greatest speed, and its energy is 
kinetic. At any intermediate place the energy is partly 
potential and partly kinetic, but the sum of the two remains 



554 APPLIED MECHANICS. 

always the same, excepting in so far as friction is wasting the 
total store. IS ow, in time-keepers the office of the mainspring 
is to give just such supplies of energy to the balance as are 
necessary to replace the loss by friction ; and we have to ask 
the question — At what part of the swing of a pendulum or 
balance can we give to it an impulse which shall increase its 
store of energy without disturbing its time of oscillation'] 
The answer is this. If a blow is given to the bob of a pendu- 
lum when it is just at its lowest point, energy is given to the 
pendulum ; we give it power to make a greater swing, but the 
time which it will take to make this greater swing is just the 
same as the time it would have taken for a smaller swing. 
This middle point is the only pomt at which we can give an 
impulse to the bob without altermg the time of its swing. In 
the lever escapement, and in other detached escapements of 
watches, the impulse is always given just at the middle of the 
swing. 

EXERCISES. 

.1. A point describes a S.H. motion of 1^ foot amplitude in a period 
of -fth. of a second. Find its maximum Telocity and maximum accelera- 
tion. Ans., 21 TT feet per second ; 294-71^ feet per second. 

2. A piston ^th. rod and crossh.ead weigh 350 lbs. If they have a 
S.H. motion with amplitude 1"1 foot, and if the maximum accelerating 
force is equal to that produced by a pressure of 15 lbs. per squai-e inch on 
a piston 14 inches diameter, what is the periodic time ? Ans., 037 sec. 

OTHEPv EXAMPLES OF PERIODIC MOTION. 

450. "When the periodic motion of a body is not simple har- 
monic, we find that by imagining the body to have two or 
more kinds of simple harmonic motion at the same time we 
can get the same result. Thus, it is known that a float, em- 
ployed to measure the rise and fall of the tide by marking 
on a moving sheet of paper with a pencil, has a motion which 
is periodic and not simple harmonic. If horizontal distances 
represent the motion of the paper (unwound from a barrel by 

means of clockwork), and 
^-^.^^ ^.^ therefore represent time, 

^N.^^ ^ ^ y^ X^ and if vertical distances 

\^^y^ \_^y^ mean the rise or fall of 

Fig. 314. water-level in feet, we get 

such a curve as is shown in 

Fig. 314. Now this is not a simple harmonic motion. The 

difference becomes evident if vou plot on souared paper the 



APPLIED MECHANICS. 



555 




Fig. 315. 



distances o A, o b, o c, o d, o e, of, etc. (Fig. 308), for 
equal intervals of time, for you will get a curve like Fig. 315, 
which is easily recognised, 
and is called a curve of sines 
or cosines. But it has been 
found that if we take certain 
curves of sines whose periodic 
times are — 1^ the semi-lunar 
day ; 2, the semi-solar day, and 
some others, their amplitudes 
and epochs being properly chosen and draw them on squared 
paper, and add their ordinates together, we get the curve 
which shows the real rise and fall of the tide. In the very 
same way we can combine simple harmonic motions to arrive 
at any periodic motion. A good way of combining simple 
harmonic motions experimentally is to let a body hang from a 
string which passes over two ^ ^_. 
or more movable, and the same 
number of fixed, pulleys. These 
pulleys are pivoted on crank 
pins, and their pivots are made 
to revolve at any desired rela- 
tive speeds, and each gives to 
the body a purely simple har- 
monic motion by its action on 
the string. The body gets a 
motion compounded of the mo- 
tions of the pulleys, and if it 
is an ink-bottle or pencil press- 
ing on the paper on a revolv- 
ing paper roller, we get a time 
curve of the periodic motion. 
This is the principle of the 
construction of Lord Kelvin's 
Tide-Predicting Machine. 
451. When a body can 



swing east and west under the 




Fig. 310. 



influence of forces which have 

no tendency to move it except 

in a direction due east and west, and if forces acting due north 

and south can make it swing in their direction, then both sets 

of forces acting together on the body will give it a motion 



556 



APPLIED MECHANICS. 



compounded of the two simpler motions. Thus, a ball a 
(Fig. 316) Ls suspended by a string, pa, which is knotted at P 
to two other strings, p s and P s', equal in length, and fastened 
at s and s'. The ball may swing in the direction E o w as if 
it were the bob of a pendulum hung directly from the ceiling 
at p', but it may also swing in the direction x o s at right 
angles to e o w, and if it does so it swings as if the point P 
were the fixed end of the pendulum a p. When it swings 
under the influence of the two sets of forces tending to make 
it move both ways at once, the moticm. of a is compounded of 
the other two simpler motions. If p a is one-quarter of the 
length o p', then the east and west swing takes twice as long 
, as the north and south 

^ II 10 9 8 7'^ swing. If p A is one-ninth 

" ' ^ - '^ ^ ^ - of o p', then the east and 

"J^^ n? °° west swing takes three times 

as long as the north and 
south swing. The motion of 
A is sometimes very beauti- 
ful, and the experiment is 
easily arranged. 

452. The motion is quite 
easily represented on paper. 
Thus, in fig. 317, a' m is the 
north and south direction, 
and A M, at right angles to 
it, is the east and west di- 
rection. Let the points 0, 
1, 2, etc., in each of these lines be found as in Fig. . 308. 
Let the bob be supposed to go from to 1 in a' m in the 
same time as it goes from to 1 in A m. Kotice that we 
have twice as many points in am as in a' m, showing a 
slower oscillation in the direction a m. We can begin to 
number our points anywhere, remembering that when the 
bob completes its range it comes back again in the opposite 
direction ISTow put marks where the east and west lines 
meet the north and south ones, drawn tlirough corresponding 
points. It is evident that the curve drawn through these 
successive marks is the real path traced out by the ball when 
acted upon simultaneously by the two sets of forces urging it 
in a north and south, and an east and west directioEL 

If we have the same number of points in a' m as in A M 




Fig. 31- 



APPLIED MECHANICS. 557 

we get a circle, ellipse, or straight line, as in c, b, a, Fig. 318. 
This represents the motion of a conical pendulum free to 
swing in every direction. Again, d, e, f, and many other 
curves that might be drawn, represent the case which we took 
up in Fig. 317, where one vibration is twice as quick as the 
other. If the time of vibration in A m is to the time of 



A „<^ 



C. 

vOOO< 



ebdoooocoo cooooooocra 



o o 



^J^^ 



o 
o o o 



vibration in a' M as 2 to 3, we get curved paths like g, i, Jj 
and so on. In experimenting with the pendulum, Fig. 316. 
it will usually be found that slight inaccuracies in the lengths 
of the cords will cause a continual change to go on in the 
shape of the path traced out by the ball. 

We can produce these motions by spiral springs, and in 
other ways. Thus, for example, if we use instead of the strip 



558 APPLIED MECHANICS. 

of steel, in Fig. 312, a combination of two strips, b and b', as 
in Fig. 319, so that the heavy bright bead A is capable of 
vibrating in two directions at the same time, we get the same 
combinations of simple harmonic motions, depending on the 
point at which b is held in the vice c. 

453. When a body has a periodic rotational motion about 
an axis like the balance of a watch or a rigid pendulujn, we must 
no longer speak of the force causing motion, and the mass of 
the body, and the distance of displacement; but if we sub- 
stitute for these terms, moments of forces, moment of inertia 







Fig. 819. 

of the body and angle of displacement, we have exactly the 
same rule for finding the periodic time of oscillation. The 
periodic time is 6-2832 times the square root of the anciilar 
displacement of the body at any instant, divided by the angular 
acceleration at that instant. And we know that angular 
acceleration may be calculated by dividing the turning moment 
acting on a body by the moment of inertia of the body. A 
point in the balance of a watch swings in circular arcs, but if 
we only take account of the distances which it passes through, 
and suppose it moved in a straight line instead of in the arc 
of a circle, the motion is very nearly simple harmonic. If 
there were no friction or other forces acting on the balance 
except the turning moment of the balance spring (see Arts. 
521-2-3), and if the moment of the spring were always exactly 
proportional to the angular displacement of the balance, the 
motion would be simple harinonic. 

"SVe shall see in Art. 522 that the turning' moment due to the 

spring is •r-o'j 0, if e is the modulus of elasticity of the spiing, b its 

breadth, t its thickness, and I its length, and if is the angular 
displacement in radians. Angular acceleration is this moment 

E bfi 6 
divided bv moment of inertia i of the balance, or z-^^, — . Hence 

12^1 

I 9 li 
angular displacement 6 divided by angular acceleration is " ^ 

so that the periodic time of the balance is t = 6-2832^ / — — "■ 



APPLIED MECHANICS. 559 

Increasing the moment of inertia of the balance or the 
length of the spring makes the vibration slow. Increasing the 
breadth and, what is still more important, increasing the 
thickness of the spring makes the vibration quick. As we shall 
see in Arts. 521-3 that our calculation of the turning moment of 
the spring is not quite right, that the dimensions of the 
balance and spring alter with temperature, and that, above all, 
the elasticity of the steel alters with temperature, and with its 
own state of fatigue, the rule is not perfectly true, nor can 
any balance be regarded as taking exactly the same time for 
its oscillation in different lengths of arc. At the same time it 
is of great help to the watchmaker to know that with con- 
siderable, although not with perfect accuracy, the time of 
vibration of a balance is proportional to the square root of the 
length of the spring, and so on. For example, suppose the 
spring is 3 inches long, and the balance makes one swing in 
0"251 second, now if he wishes it to make a swing in 0'25 
second, he must shorten it in the ratio of -251 x '251 to 
•25 X -25, or in the ratio -063001 to 0625, so that the length 
of his spring ought to be 3 x -0625 -f -063001, or 2-976 
inches — that is, it ought to be shortened -024 inch. In the 
same way he can calculate the effect of adding little masses at 
any distances from the centre of the balance, so that its 
moment of inertia may be increased, and the balance made 
slower in its swing. The same law tells him how he can 
compensate the balance, so that when in summer the steel of 
the spring loses its elasticity, some of the mass of the balance 
will come nearer the centre, in order that the moment of 
inertia may diminish in the same proportion. 

454. Compound Pendulum. — The simple pendulum described 
in Art. 446 is not like the pendulums used in practice. In these 
the bob is not so small that we can consider it as a point ; the 
long part is not a thread but a stiff rod of metal or wood, and 
there is usually a knife-edge for support, about which it can 
turn with little friction. In common clocks, however, the top 
end of the pendulum is a thin strip of steel held firmly in the 
chops, but the easy bending of this strip is such that we may 
imagine an equivalent pendulum to move freely about an axis. 
Employing our general rule of Art. 453, we find how to calculate 
the time of vibration. 'J'his compound pendulum vibrates in 
the same time as a certain simple pendulum, called the equi- 
valent simple pendulum, whose length we find by experiment. 



560 



APFLIED MFjCHASIGB. 



In Fig. 320 kt a be die axis of swugpcnsitm, G ilie centre 
o€ grsvi^, and P a point in llie continiiation of tlie line 
8 6 sach that s p is the length of the eq[iiiTalent simple 
psidnlom. Th^i p is called the oentn ^ 
asaBatitm, and it is also known to be the 
eemire of 'percumion, of the p^idoliun (90s 
Artu 402). It can be proTed that if the 
pendnlum be inverted and made to Tibfate 
about a paralkj axis Uuoogfa p, it will 
Tibiate in exaetty tiie same time as it does 
aboot 8 ; md it was in this waj, by in- 
verting a pendnlnm which had two knife- 
edges, and adjosting tiiese nntil the pendolnm 
took the same time to vilKate aboat one 
as about the odier, and dien measoiing 
the distance between, tiieni, ^lat Gaptain 
Kater found the laigtii dt tiie sim|de pen- 
dnlnm wiiich Tibrates in a giTen tone. 
Hiis method is still ^nj^oyed in gnmtatian 
ex per i iHCTitB everpdiexe to find die value 
of g, which is 32-2 foet per second per seccMid at LcMid<m. 
It is an exeellrait labcNcatGcy ex^mscL 

If s ie file asds of voBpsadsm, e fiie eentze of g i a * iij , w Hie 
weig^ of flie pendnbiB, Sicb. flie iiMiwiiKMit wiSb 'vki^ gssiity 

trrr^ i?ie t*'^'^'"^'""" te re-' '-irn to its pnaaftinw of test is w x c sr ; 
- 1" li '.—-. ":.~ --T = 't i-T^BSDzed. in zaoians^ and ii it n \&y 
\z. 11 T.":!:- i_i;iit-- :.- ;'_i_:-: rXi^Qyeq^La! k' "wr x se x angles o. 
r^T 1. : -. : _ z_ 1= : "ir^Ti :~ i ^iing; fln« "by 1^ flie 

. = .-.:/— --^^-^— 




N^;^ 



. . (1)- 



6-2832 






APPLIED MECHANICS. - 561 

where y is 32 "2. In the simple pendulum sk and sg are equal, 
and (2) gives the same rule which is given in Art. 446. However, 
in an ordinary pendulum, s k and s g arc not equal, but s e.^ -j- s o 
is equal to some length such as s p, and our rule becomes 



^/ 



6-2832 / -- ,oi 

- . . . . (3). 



Evidently ev is the length of the imaginary simple pendulum ivhich 
mould vibrate in the same time as our real jiendulum. The imaginary 
point p has been called the centre of oscillation, because when the 
pendulum is inverted and made to vibrate about an axis through p 
it vibrates in the same time as before. 

To prove this it is necessary to return to equation (1). We 
know that i is equal to the moment of inertia of the body calculated 
as if all its mass existed at g, together with the moment of inertia 
of the body as it is at present, but calculated about an axis through 
G parallel to the present axis ; that is, 

I = — s g2 + - ^2 
9 9 

where k is some length unknown to us just now, being the radius 
of gyration about the axis through the centre of gravity. Eule (1) 
becomes 



T = 6-2832 



ov T = 6'2832 



V 
V 



:^SG2+^;fc2 



Ic^ 



That is, the length of the simple pendulum which will ^dbrate in 

^2 

the same time ig s g + — , and we have already found it to be s p 

S G 

in equation (3) ; so that gp = — , or g p x s g = F, But in the 

very same way, if we considered the pendulum as ^dbrating about 
p, we should find the length of the equivalent simple pendulum to 

^2 

be greater than g p by an amount equal to — , and we know that 

s G is equal to this amount ; so that s p would, as before, be the 
length of the equivalent simple pendulum. The axes of oscillation 
and SKsjjension are therefore i) iter changeable. 

455. Exa7nples. — The bar of Fig. 321 with two adjust- 
able masses may be fixed to one end of a wire, the other end 
of which is fixed to the ceiling. By twisting and untwisting 
the wire the bar will oscillate with a motion which is much 
more nearly simple harmonic than that of the balance of a 
watch. Students who experiment with such a bar can adjust 



562 



APPLIED MECHAXICa. 



lie weights A and b at anj distance from tlie axis (there ought 
to he an engraved scale on the har), so diat the nunnent of 
inertia can he vari»l. They can fastoi Uie har at tibe end of 
a wire, or lUiey can use it as in Fig. 321, with a flat spiral 




F^ SSI. 

spring, or as in Hg. 322, with a cylindric spiral ^ring; and 
tiie rate erf its vibrataMi gives <me of the hest ways of in- 
vestigating ihe twisting mcHnoits of wires and snch springy 
when strained l^aoa^ given an^es. 

In ^e ease of a wire the twist always tends to hring the 
har to its position of rest with a moment whieh is proportional 
to ike angle of displacememt from this posUwnr— it is this 
nrcmefty whidi caoses the motion to he simple haimonic 
This moment is abo proportional to the fnarth power of the 



APPLIED MECHANICS. 



.63 



diameter of the wire, and it becomes less as the length of tla 
wire is increased. By means of a circular scale and a pointer 
we can measure the extent of each swing, and this is found to 
decrease gradually, due to friction with the air and the in- 
ternal friction or viscosity of the metal. The amount of 




diminution of swing gives us a means of determining the 
viscosity, and the apparatus can so easily be fitted up that no 
person y>^1io wishes to understand the properties of materials 
can be excused from making these experiments. This is a 
common method of finding n, the modulus of rigidity of a 
material. 

If the leng-th of the wire is I inches, its diameter d, and if n is 
its modulus of rigidity {see Table XX.), then from Art. 295 we see 
that the moment with which the wire ants on the bar. when its 



56 i APPLIED MECHANICS. 



angle is 6 from the position of rest, is - n t d^. If the naoment of 

o'l i 

inertia of the bar is i (we are neglecting the fact that the wire 
itself has some mass which has to he set in motion), then the 
moment, divided by i, is the angular acceleration ; and using this 
quotient as denominator, and 9 as numerator, extracting the square 
root, and multiplying by 6-2832. or 2 tt. by the general rule of Art. 
453, we find the square of the period of a complete oscillation to 

be T*^ = 5j— • E we are in doubt as to our calculated i, we can 

find it experimentally ; and this is very necessary in many magnetic 
exj)erim.ents. "We add a known moment of inertia i (say two equal 
small masses at equal distances from the axis), and find the new t 
(call it Tj). Then Ti-/t^ = (i + ^)h, and. i may be found. 

When motion is slow, the friction in fluids is proportional to 
the velocity, and any friction which follows this law is called fluid 
friction. A great many -vibrating bodies tend to come to rest by 
the action of such friction as this ; and it is found that if the 
friction is numerically / times the angular velocity, then the 
logarithm of the ratio of the length of one complete swing to the 
next is nearly equal to k times the periodic time. Hence this 
logarithmic decrement, as it is called, is proportional to the friction 
co-efficient. If we obserA'e twenty-one elongations on one side of 
the middle position, then one- twentieth of the logarithm of the first 
elongation divided by the last is k times the periodic time of 
oscillation. 



EXERCISES. 

1. Bifilar Suspension. — In many meastiTing instruments a body is 
suspended by two thin wires nearly vertical. If the vertical length of 
each of these is I, the distance between their ends at the top a, and at the 
bottom b, and the weight of the body w, it is easy to show that for a 
small angular displacement 6, the moment tending to bring the body to 
its position of rest is very nearly (neglecting torsion of the wires them- 
selves) ^ — w6. Fiud the time of vibration of such a body when its 

moment of inertia is known. In truth, the constraiuing moment is 
proportional to sin. 6. 

2. A magnet, turning on a frictionless pivot at its centre of gravity, is 
subjected to a turning moment h sia. 0, or very nearly h 9, due to the 
earth's magnetic action, if it makes only a small angle 6 with its position 
of rest. Find the time of a -vibration if the moment of inertia is known, 
and show that the square of the time of -sibration of the magnet in 
different places is inversely proportional to h. Fiad what is the effect of 
adding a known moment of inertia, and show that the observations on the 



APPLIED MECHANICS. 565 

two times of vibration enable us to calculate the original moment of 
inertia if it was unknown. 

3. Prove that the time of complete oscillation of a ship is 27r^/ \/ gd^ 
where d is the distance from the centre of gravity to the metacentre. 

456. Stilling of Vibrations. — When a simple harmonic 
motion is represented on paper in the manner described in 
Art. 450, we have a curve of sines. The curve may be ob- 
tained by producing the lines b b', c c', etc., of Fig. 308, 
cutting them at right angles by equidistant horizontal lines, 
and joining the successive points of intersection so found. It 
may also be drawn by finding from a book of tables the sines 
of 0°, 10°, 20°, etc., and plotting and sin. 0°, 1 and sin. 10°, 
2 and sin. 20°, etc., on a sheet of squared paper. 

A curve of sines expresses the fact that, if d represents the 
displacement of a vibrating body from its middle position after an 
interval of t seconds since it was at the middle of its course, then 
d = a sin. 2>^ where a is the greatest displacement of the body 
fi-om its middle position. This displacement is usually called the 
amplitude of the vibration. If T is the time of a complete vibration, 

it is easy to see that the equation \s, d = a wa.. — t, or a sin. 27r/' . ^ 

if/ is the frequency or number of complete oscillat'Cl>'- ^er second. 

If we make the bob of a pendulum terminata below, in a 
tube which can act as a pencil-holder, and in which a well- 
fitting pencil can slide freely, and if we move a sheet of paper 
at a uniform rate underneath this pencil at right angles to the 
direction of motion of the pencil, a curve of sines will be traced 
out, if the pendulum swings without friction. But in practice 
we always find that, what with the friction at the point of 
support, friction with the atmosphere, etc., a pendulum's 
swings get smaller and smaller — that is, the amplitude of 
the vibration gets less and less as time goes on, until the 
pendulum at length comes to rest. 

This motion is not a simple harmonic motion, but, wdthin 
certain limits, each swing may be regarded as very nearly a simple 
harmonic motion. Practical men who deal with oscillating bodies, 
such as pendulums, ships, tuning forks, magnetic needles, and 
suspended coils of wire, usually assume that the motion during 
each swing is a simple harmonic motion. The frictional resistance 
to motion of any ordinary "S'ibrating body in a fluid mediuia, or of 
a magnetic needle vibrating neai- any body capable of conducting 
electricity, is almost always such that the quicker the motion tli^ 



566 APPLIED MECHANICS. 

greater the friction {see Art. 64) — that is, frictional resistance is 
proportional to speed ; and in this case it is not difficiilt to show 

2 IT 

that, instead of the law d = a sin. t . . . . (1), we have the law 

d = «e~"^ sin. — t . . . . (2). That is, if the strength of the spring, 

or other governor of Titration, and the character of the vihrating 
hody are such that without friction the law would he (1), then, 
when the vihration is damped by frictional resistance of the ahove 
character, the law of the motion "becomes that given by equation (2). 
Here a is a constant which depends on the chai-acter of the friction. 
Thus a is greater when a pendulum swings in water than when it 
s^ving•s in air. Also, t^, the periodic time of the vibration, is no 
longer the same t as it was for undamped vibir.li ns, and the 

: jlation between t and ti is -^ = — s + -: — s .... (3) ; or, if / is 
the imdamped frequency, and f^ the damped frequency, then 

2 

/2 =/i2 + — ^ .... (4). In order to get exact ideas on this 

subject of f\xi damping of vibrations, the student ought to plot on 
squared jviper a curve such as o a' b' c' d' e' f' g' h' i, Fig. 323, 
which coriesponds with equation (2). Thus, let us suppose that a 
body undamped in its vibrations gets an impulse which sends it 
from its position of rest in such a way that its amplitude is 10 
inches, and let the time of a complete oscillation be 1'6 second. 

6'2S32 
Then the law of its motion would be a; = 10 sin. ~ ' t, or 

I'D 

« = 10 sin. 3-927 t . . . . (5) where a; is in inches, t in seconds, and 
the angle 3-927^ in radians.* 

If, now, the friction is such that a = 07, we find from (3) that 
the time of an oscillation is practically unchanged. Find, there- 
fore, the original curve of sines by calculating the second column 
of the following table. The numbers of the first two columns 
plotted on squared paper would represent the undamped vibrations. 
But for damped vibrations the numbers of the second column have 
all to be mrdtiplied by e"^'"^^ ; and if we denote this multiplier by 
the letter s, we see that z being e"^*''^, or log. z = -0-304;'. We 
have calculated z for the various values of t, and placed the results 
in the third column. Multiplying, therefore, the respective 
numbers of the second and third columns together, we get the 
fourth column of numbers ; and plotting the numbers of the first 
and fourth columns on squared paper, we find the curve which 
shows the nature of the damped vibrations. 

* We may ^vl•ite (5) in the form cc = 10 sin. 225*. In this case the angle 
225t is expressed iii degrees. 



568 



APPLIED MECHANICS. 





10 siu. 3-927 t, or 10 






t in seconds. 


sin. 225 1, if angle is 
taken in degrees. 


,-0-7t 


e-0'^hosin. 3^9-27< 








1 





0-2 


7-07 


•869 


6^14 


0-4 


10 


•756 


7^56 


0-6 


7-07 


•657 


4^65 


0-S 





•571 





1-0 


- 7-07 


•497 


- 3-51 


1-2 


- 10 


•432 


-4^32 


L-4 


- 7-07 


•375 


- 2-65 


1-6 





•326 





2-0 


10 


•247 


2^47 


2-4 





■186 





2-8 


- 10 


•141 


141 


3-2 





•106 





3-6 


10 


•080 


•8 


4-0 





•061 





4-4 


- 10 


•046 


- 46 


4-8 





•035 J 





6-2 


10 


•026 


•26 


5-6 





•020 





6-0 


- 10 


•015 


- 15 


6-4 





•Oil 






457. Some students may find it as instructive to first draw a 
curve of sines, then draw the logarithmic curve, correspond- 
ing to column three, on the same sheet of squared paper, and 
multiply the ordinate of one curve by that of the other to get 
the ordinate of the real curve which exhibits the damped 
vibrational motion. This is what has been done in Fig. .323 ; 
OABCDEFGHI is the curvB of sines, L P Q is the logarithmic 
curve, showing how rapidly the amplitude of the vibration 
diminishes, and o a b c' d' e' f' g' h' I is the curve which repre- 
sents the actual motion of the vibrating body. In this figure 
the logarithmic curve is drawn to such a scale as seemed con- 
venient for showing its properties distinctly. It would be 
very easy to dilate on the nature of the resulting curve o a' b', 
etc., but this book is written to help students who are earnest 
enough to calculate the above numbers and plot the curve, and 
when they perform these operations they will have very clear 
notions about the motion we have been investigating. 

Exercise. — A heavy disc, suspended by a wire, vibrates in each 
of a number of fluid media, its periodic time of vibration in all 



APPLIED MECHANICS. 569 

being sensibly the same, or 1-5 second. The ratio of the amplitudes 
of two successive swings in one dii-ection being 0-9 in one fluid— 
that is, the second swing beiug only iiine-tenths of the first, and 
the third being only nine-tenths of the second, and so on — and 0-8 
in another fluid, and 0'7 in another, what numbers will express the 
relative viscosities of these fluids ? 

Here we have, taking common logarithms, 09 = g-l"oa £,^^. ^.j^g 

— loo-. 9 
first fluid, so that - log. 0*9 = 1-5 a log. e, or a =; '^' — , that 

i 'O log". 6 

is, a =: 0-07. In the same way a = 0-15 and a = 0-24 for the 
other fluids, and hence 7, 15, and 24 are the required numbers 
expressing the relative viscosities as measured by the ^abrating disc 
method. 

A very slowly swinging disc and pointer will enable iis to 
plot the complete curve from actual observations. The nature 
of the motion when the frictio]i is that of solids rubbinof on 
solids is studied in my book on the Calculus. 



570 



CHAPTER XXYT. 



MECHANISM. 

458. In Art. 179 we pointed out liow skeleton drawings and 
models may be made useful in cases where velocitr ratios vavj greatly. 
TTe shall now giye a sketch or outline of a general theory from 
which students may find benefit if they fill it in for themselyes. 
We cannot say that there is as yet any satisfactory treatise on this 
subject. The most interesting part of it (to ns) is that which 
concerns the mechanism of steam and other engines, and this will 
be found in the author's book on the Steam Engine. 

459. We haye already referred to spur and beyil gearing used to 
driye one shaft from another at uniform yelocity ratios. Consider 
at any instant two teeth in contact ; each of them is a rigid piece 
rotating about a fixed centre and acting on the other by rubbing 
contact. We shall now briefly refer to this way of tracsmitting 

motion. 

Let A w be a body 
moATng in the plane of the 
paper about the axis a, let 
B y be another body moying 
about the axis b, and let 
these TWO bodies keep in 
contact, as we see_ them in 
contact at i . Xow, this 
keeping in contact, what 
does it mean ? It is that, 
whateyer may be the rela- 
tive tangential or sliding 
motion at the point of con- 
tact p, they haye the same 
yelocity in the dii-ection of 
their common normal there 
at the instant when we 
study them. Consider their 
motions during an exceed- 
ingly short interval of time 
St. There are tsyo points p 
to be considered. One is on 
the body a w, and it moyes to q, where p a is an arc of a cii-cle about 
A as centre ; the other p is on the body b y, and it moyes to k, 
where p k is an arc of a circle about a as centre. The angular 
velocity- of A w being a, the angle p a q is a . St : the angular 
yelocity of b y being b, the angle -bbu is b . St. As St is con- 
sidered smaller and smaller, p a and p r may be considered more 
and more neai'ly short straight lines. i? o, = at? . a . St and 
p B. = B p . i , 5^. Now obserye that if t p s is the direction of 




APPLIED MECHANICS. 671 

the common normal to tlie two bodies at p, the normal component 
p s of the motion p q must be the same as that of the motion p b; 
in fact, the straight line o. s r must be at right angles to the 
normal t p s. Now draw t e parallel to a p, and observe that the 
two triangles t e p and q p r have their sides o, r and t p, r p and 
p E, p a and E T respectively at right angles to one another, and 
hence they are similar ; that is, 





QB 


:rp:pq=tp:pe:et... 


.(1), 












or !£ - l^. 
















P Q E T 














Hence ^^"^ ^^ or- 


AP . 


P E 












AV . a E T' a 


~ BP . 


ET 








But because 


1 TE 


is a line drawn parallel to ap, a 


side 


of 


the 


triangle a p b, we 


have 
















AP AB P E 


AT. 














— — — and — — 


■ » 








* 






ET BT BP 


AB 














therefore ^ ^ • ^ ^ 


AT 














ET . B p ■ 


~ BT 














Hence ^ - ^ . . . 


.(2). 














a BT 












Hence the 


ratio of the angular velocities is 


inversely 


as 


the 



segments into which the common normal at the point of contact 
divides the line of centres. Hence, if the ratio of the angular 
velocities is constant, the common normal at the point of contact 
p passes always through the same point t in the line of centres a b. 

O "R. "P T "R T 

Notice also that — = — . But p q -^ <? . ap . S^!, t e = ap — , 

P Q T E A3 

Hence it will he found that 



AB a -V h 

Q K = P T (« + J) S?;. 

That is, the slipping speed at p is the speed at the end of a radius 
T p when the radius revolves at the angular velocity a ■\- h. Hence 
there is always slipping at p, unless p is on the line of centres and 
the speed of slipping is proportional to the distance p t. 

The student will find that when there is friction, if d p t = 
d' p T = angle of repose, then p d is the direction of the mutual 
force when aw rotates as shown, with the hands of a watch; 
whereas p d' is the direction of the mutual force when a w rotates 
in the opposite direction, so that b v is the driver. 

If we have given the shape of b v p, and we desire to find the 
shape of a piece a w p which will gear with it at a constant angular 
velocity ratio, make a template of bvp, and arrange that when 
this template is moved about the fixed centre b a sheet of paper 
shall move about a through the proper angular distances. If for 
each position of the two the curved shape of b' p v' be drawn on 
the paper, the pencil marks will show the proj)er shape of a w p 
if it is to touch b p v. In fact, a w p will be the envelope of the 
shapes of b' p v'. "V\Tien b' p v' is circular with b as centre, a w p 
will also be circular. True rolling will be possible, for p will be 
at T, and as t will be constant in position the angular velocity 



572 APPLIED MECHANICS. 

ratio will be constant. When b' p v' is the arc of an ellipse whose 
focus is at B, and if the distance b A is equal to the major axis of 
the ellipse, awp will he a similar ellipse; there will he true 
rolling, with changing angular velocity ratio. The other foci may 
he connected by a link. If b' p v' is shaped like an equiangular 
spii-al whose pole is at b, AWP-will he a similar spiival, and there 
will he true rolling between them, but with changing angulai 
velocity ratio ; in this way lobed wlieels are formed to gear together. 

460. Discs or cylinders touching each other, their axes parallel, 
are used for friction gearing. If the horse-power h is to be trans- 
mitted, and V is the common circimiferential speed in feet per 
mimite, p being the necessary tangential force, p = 33,000 h/v. 
Slipping is to be impossible, and therefore p -^ ^ is the force 
necessary to press the cylinders together. If this force acts 
through the bearings of the two shafts, it is usually found in 
practice that, unless at very high speeds and with small power, 
there is so much practical difficulty that the gear is never used. 
Compressed paper and leather have been used to work with iron. 
Sometimes nest-gears are used to produce the necessary pressure, 
iLtut when the necessary pressure has been produced it has led to 
disintegration of the surface, or such local elastic changes of shape 
as produce annoying sound. In one Ci^se. where. the driven pulley 
is very heavy and the pressure is produced by its own weight and 
that of the spindle and part of the weight of the rotating armature 
of a little dynamo machine, the gear has been used satisfactorily. 
Wedge-shaped grooves and projections have been cut in the rims 
of the pulleys, and sufl&cient grip has been produced in this way, 
but there is no longer true rolling. 

461. ^A'hen we attempt by using teeth to get the necessary driv- 
ing forces we introduce sliding contact, using spur, bevil, and skew 
bevil wheels ; the names pitch circles, pitch cones, and pitch 
hyperboloids being used for the friction gear, which would run 
with the same velocity ratios. When the axes are not in one 
plane, frusta of hyperboloids, generated by the rotation of the 
same straight line round both axes, will gear with one another, 
always touching along a straight line. But there will not be 
simple rolling ; there is sliding along the line of contact. 

Every student ought to study the shapes of spur-wheel teeth ; 
it is easy to apply one's knowledge to other kinds of teeth and 
rubbing and rolling gear. Nothing illustrates the fact that we do 
not really think, so well as this, that all the principles for the 
proper construction of worm-wheel teeth and chain gearing were 
to be found in books many years before there existed any good 
worm-wheel teeth or chain gearing. The subject, like all other 
parts of machine design, is best studied when one diraws things to 
scale, and there are now many books to assist the student in 
machine design. Perhaps it will be well to neglect almost all the 
mathematical parts of such books on strength. 

462. Suppose we have two curved rollers, v t v moving about 
the axis b and w t w aboxit the axis a, and suppose that these are 
capable of rolling on one another as they rotate ; they touch at 
T, and the anoular velocities about a and b are as b T to a t. Now 



APPLIED MnCIIANlCS. 



573 



suppose we wish wheels, with teeth centred at b and a, to have 
exactly tlic same angular velocity ratios as the rollers ; if k' r v' 
and a' r w' are the shapes of the teeth in contact at i', it is 
necessary that during- the motion the common normal at p should 
pass through the j^itch point t. To effect this object it is necessaiy 
that ](' p v' and a' p w' 
should be two troch- 
oidal curves, geneiated 
by the rolling of a 
ciu've inside the curve 
T V and outside the 
curve t w. Thus in 
Fig. 325 imagine the 
cui^'c V and the curve 
w to move about b and 
A, rolling on one an- 
other and keeping in 
contact at t in the 
straight line b a, and 
imagine the curve p p't 
to roll also, keeping in 
contact with both v 
and w at t. It is really 
rolling inside v and 
outside w. If any point 
of p p' T, such as p, is 
always on the contours 
of the two teeth, the 
straight line p t is al- 
ways normal to these 
teeth at p, their point 
of contact, and we have 
ensured that the com- 
mon normal to both 
passes through the 

pitch point t. w and y may be ellipses, but they are generally 
circles, a and b being their centres. If p p' t is a circle, rolling- 
inside V and outside w, the trochoidal curves are called hypo- and 
epi-cycloids. When a number of wheels arc to gear any one with 
any other, we usually choose one rolling cii'cle for the insicles and 
outsides of all the pitch cii-cles ; it is taken of a diameter equal to 
the radius of the smallest pitch circle. When it rolls inside this 
smallest x^itch circle the hypocycloid is a radial line. A rack may 
be regarded as part of a wheel of infinite diameter. Sometimes the 
trochoidal cm-ves are involutes of circles, the rolling curve being 
really a straight line or an infinite circle. Let w and v (Fig. 32()) be 
the pitch circles. Draw any line c D through t, the pitch point, and 
describe circles with b and a as centres touching this line at d and 
c respectively. Draw through t' gtii, the involute of the circle 
D G, and F T E, the involute of the circle c f. If g h and e f be the 
contours of teeth rotating about b and a, their common normal 
remains always c d. Thus at any point t' in c d, if we di\aw 




Fig. 32a. 



574 



APPLIED MFCHAXICS. 



mroliites to the two circles, they must touch thrre, because a 
tangent to a circle cuts all the involutes at right angles. It is 
BA-ident that a pair of wheels with involute teeth may have less or 
more distance "between their centres without any alteration of their 
velocity ratio, this being b t : a t or b d : a c. If there is no friction, 
c D is the directicin of the driving force. It is usually more oblique 
to the line of centres in wheels with involute teeth than in cycloidal 
teeth. For the most accurate work the actual trochoidal curves 
are drawn by the rolling of templates, but there are well-known 
drawing-office rules by which we approximate to the true best 
curves by means of arcs of circles. 

463. in Art. 462 we showed how, when given the shape of a tooth, 
to find the shape of another tooth to gear with it with constant 
velocity ratio. A good example of the application of this rule is 
found when designing the shape of the tooth of a worm wheeL 




The shape of the thread of the worm being chosen, any section of 
the worai thread by a plane at right angles to the axis of the worm 
wheel may be diawn by elementary practical geometry. Regard 
it as the tooth of a rack. The section there of the tooth of _ the 
worm wheel must be the shape of a tooth to gear with the given 



APPLIED MECHANICS. 575 

rack tooth, and must "be di-awn by some sucli nile as we have given 
in Art. 462. 

464. Sometimes pieces centred at a and u, acting upon one 
another, do not act directly as shown in Fig. 324, but through the 

\ 

8 




agency of a third body. Thus, if pins on v and w are connected by a 
link w V, and if the pins are frictionless, we know that the direction 
of the force at any instant must be in the direction of the ( entres of 
the pins or in the line w v. It is easy to show as before that if a 
is the angular velocity of aw and h the angular velocity of bv, 

then V = — , if T is where w v cuts the line of centres a b. The 
h ta' 

complete study of the relative motion of four links such as aw, 
w v, V B, and b a is of course a very complicated business if we 
imagine them to be of all sorts of lengths. We may, if we please, 
imagine any one of them fixed and consider the motions of the 
others. Thus in Fig. 328 consider a b to be fixed. Think that 
A and B are merely pins in some fixed object of any shape whatso- 
ever. Now consider the other pieces to be of any curious shapes and 
lengths. The motion is taken to be in the plane of the pajper. w v is 
only a straight line joining the centres of the two pins w and v ; 
but imagine w v to rej)resent any curiously shaped body — we might 
wish to know the motion of any point in this body. The student's 
great aim is to get a correct mental picture, and if he recollects 
that Av is moving about a as an instantaneous centre or at right 
angles to a w, and v is moving about b, the whole body w v must 
just at the present instant be moving about the point c as a centre. 
We have produced av a and v b to meet at c, the instantaneous 
centre of motion of the whole body w v. One has then a mental 
picture of the motion of any point whatsoever in the body w a*. 

465. General motion of a body parallel to a plane. — Let us 
simply say a plane figure or body in its own plane. If we consider 



576 



APPLIED MECHA.NICS. 



the motion of any two points, this settles the motion of every other 
point, so we shall speak only of two points. It is easy to prove 
that when we consider two positions of the body there is one point 
of the body whose position is the same. Thus, if w v is one 
position and w' v' is another j)osition, bisect w w' by a line at right 
iingies to it, and let it meet the rectangular bisector of v v' in c. 
Evidently c is a point in the body which has not altered its 
position ; it is the instantaneous centre of the motion. Notice 




Fig. 329. 



Fig. 330 



that if w w' is parallel to v v', c is at an infinite distance ; we then 
say that the motion is one of translation merely. In Fig. 329 we 
have a piece, w v, whose ends move in the paths w' w and v' v. At 
any instant, if we draw w c and v c, normals to the paths, we find 
c the instantaneous centre. 

Consider any point p in wv, and join p c. p traces out some 
path during the motion. Notice that p c is the normal to this 
path, and a line through p at right angles to p c is the tangent to 
it. If we want to consider the enA^elope of the straight line w v, 
notice that the foot of the perpendicular from c ujDon w v is a point 
in that envelope, because it is the only point of w v which moves in 
the direction of the length of w v. The student must not imagine 
that because at the instant every point of a body is moving about c, 
therefore c is the centre of curvature of the path of w, or of v, or of 
p. c is only for the instant the centre of motion, and such lines as 
w c, V c, and p c are the directions of the normals to the actual 
j)aths of these three points. As the figure moves in its own plane 
from one position to others successively, let Ci c^ C3, etc., bo the 
successive points of thefr/ure about which the rotations take place, 
and let cic.-,c.^, etc., be the positions of these points on the fixed plane 
when each is the instantaneous centre of rotation. Then the figure 
rotates about Ci (or Ci, which coincides with it) till c^ coincides with 
('., ; then about c^ till C3 coincides with c.^, and so on. Hence, if we 
join Ci C2C3, etc., in the plane of the figure, and CiC-iC-^^ etc., in the 
fixed plane, the motion will be the same as if the polygon Ci Cg C3, 
etc., rolled upon the fixed polygon c\ c.^c.^, etc. By supposing the 
succe.-<.si\'e displacements smaller and smaller, Ave have cur^'cs, aiid 
hence any motion whatever of a plane figiu-o in its owti x>lane may 
be imagined as produced by the rolling of a curve fixed to the 
figure upon a curve fixed to the i)lane. It immediately follows 
<hat any displacement of a rigid soHd parallel to a plane may be 
produced by the rolling of a cylinder fixed in the solid on another 



APPLIED MECHANICS. 



577 



cylinder fixed in si^ace, the generating lines of the cylinders being 
at right angles to the plane. 

466. What we have said about motion in a plane is true about 
motion of figiu^es shaped to fit a spherical siu'face ; straight lines in 
the one case corresponding to great circles of the sphere in the 
other case. It is easy to have a mental image of this. Now 




Fig. 331 



imagine that any point of the superficial spherical figui'e is joined 
to the centre of the sphere, and we see that (1) if a rigid body has 
one point fixed (imagine this the centre of an imaginary sj^herical 
surface), however it may move, in any two positions of the body 
there is one line of the body which is common to the two positions. 
(2) Any motion may be regarded as due to the rolling of a cone 
fixed in the body upon a cone fixed in space. What we have said, 
therefore, about the pieces aw and bv of Fig. 327 moving about 
axes at right angles to the plane of motion may be at once applied 
to pieces moving in a spherical surface about axes meeting in the 
centre of the sphere. Or the cylindric pieces a W and b v may be 
imagined to be conical pieces moving about axes a and b, meeting 
at the vertex of the cones. Hence all that we have said alDout spur 
wheels is at once applicable to bevil wheels, which are suitable for 
shafts whose centre lines or axes meet at a point. 

467. Fig. 329 is worth a very great deal of consideration. There 
is a link w v, and motion of a point w in it is known. The shape of 
the path of v is known, to find the motion of the whole link, and of 
any point in it, at every instant. When the -paihs of w and v are 
arcs of circles, we ha^'e the four-bar kinematic chain of Fig. 327. 
When w (a crank pin) moves in the arc of a circle, and v is a block 
moving in a straight slot, one particular case is called a slider 
crank chain {see Fig. 332). When both w and v are blocks moving 
in straight slots at right angles to one another, we have the 
ordinary trammels for describing an ellipse. We have the mathe- 
matical basis of the elliptic chuck, for we may imagine any part of 
a mechanism to be at rest, and tnen the relative motions of the 
others become absolute motions. The theory, then, of Art. 464 is 
the same for a very great many mechaniems. 




578 APPLIED MECHA>'ICS. 

463. Li Fig. 327, if we imagine b a to be fixed, b in the position 
in -sv-hich it is in tte figure, w a and b a infinitely long, so that Va 
arc of a circle is really a straiglit line, we have the usual crank b v 

and connecting rod v w, 
which we can indicate 
in its simplest form by 
Fig. 332. Now take 
this as it stands. A 
fig. 332. piece b a X along which 

w slides, and the con- 
necting-rod w V and the crank v b ; think merely of the relative 
motion of each piece to the rest. Ton may imagine any one of 
the pieces fixed, and you may consider the motion of any point 
in any piece. If bx is fixed, y describes a circle, w a straight 
line, and any point in w y describes a cutyc which is some- 
what like an ellipse, only blunter at one end than the other. 
Now imagine b y = y w : y will not make a rotation. Imagine 
w Y extended beyond y as far again : that point will describe a 
straight line, and we have a parallel motion. A student must try 
these things with models. Now imagine b y fixed, and let yw turn 
uniformly. [Draw the fixed part of any curious shape, the frame 
of a machine, y and b here being two pins.] We get a quick 
return mechanism for shaping and other machines. Imagine y w 
fixed, and we have the mechanism of oscillating cylinder engines 
and of other machines as well. Now imagine w fixed, and guiding 
B X, so that it shall only move in the direction of its lengih, and 
we have a weU-known pump mechanism. ^Tien we consider that, 
even fi-om Fig. 332, with the motion of w in a line with b, we have 
obtained a number of different -looking mechanisms, and that these 
can be varied very curiously by taking various lengths of the parts, 
it T\'ill be seen that there is a nearly endless variety of forms 
deriA-able from the mechanism shown in Fig. 327. Now imagine a 
pin T in the piece w v linked to z, a pin in an. arm z b, and we have 
a much more complicated problem to study in the relative motion 
of six pieces. Any one of them may be imagined fixed, and they 
may be of aU sorts of lengths. 

469. The following example is of geometrical interest. In 
the Peaucellier cell 'Fig. 333), ab and aw of Fig. 327 are made 



Fig. 333. Fig. 334. 

equal in length, and so are by and vw. Two pieces, wc and C5. 
equal to b v and v w, are added. It is easy to see that a c v is 
always a straight hne. Also the positions of a, c, and v are such 



APPLIED MECHANICS. 



579 



that AC. AV remains constant. For if acv and wb be joined, 
and they meet in o, the centre of the rhombus, a av^ = a o^ + o w^, 
and V v>^'^ = y o- + o v^^. Hence 

A W^ - V W^ = A O^ - V O^ = (a O + V O) (a O - to) = A V . A C. 

Hence, if a is fixed, and c traces out any curve, v -ttdll trace out 
the reciprocal curve. If c is constrained by a link o' c to move in 
a circle, and if o' a = o' c, A" Avill describe the reciprocal curA^e, 
which in this case is a straight line. If o' c is not equal to o' a, v 
will describe the arc of a circle of much greater radius than c's, 
470. A'NTien the four links of Fig. 327 form a pavaUelogi am, as 




Fig. 335. 



A w v B of Fig. 331, then, if p is a point anyA\-here rigidly part of the 
link vr \, but in the line w a', and if at any time it is in the same 
straight Hne Avith a, and a point q, AN'hich is rigidly a part of the 
link BA-, and in the line ba^, then p and q will ahvays be in a 
straight line with A. We can imagine now that either q or p or 
A is a fixed point, and the other two aa^II follow j)aths which are 
similar to one another. 

Thus, for example, if aavvb (Fig. 335) is the four-bar 
mechanism of Fig. 327 (a b need not be di^awn because we imagine 
it fixed), and if a w d is one piece, a^ v and p d being links equal to 
w D and w a^ respectively, then the pin p will travel in a path 
similar to that of q if q w : A aa' : : p D : a D. If q traA^els A^ery 
nearly in a straight line (if b v : a w : : Q w : q v, we haA^e a 
near approach to straight-line motion of q, so long as b v and a w 
do not make too great an angle with one another), then p will also 
traA-el A-ery nearly in a straight Hne. 

471. Again, if a w v b (Fig. 336) is a imrallelogram, and if p is a 
point rigidly part of the link b a', and if q is a point rigidly part of the 
link w A'-, and if p and Q are so placed that the ratios p a' : a^ b : b p are 
the same as the ratios v q : a w : w v, then it can be shoAvn that a q 
and if p and a are so placed that the ratios p y : y b : b p are the 
same as the ratios a'Q : qav : av a', then it can be shown that aq 
and AP always make the same augles with one another, and that 
the ratio of the lengths a q and a p keeps constant. In the 
triangles p v b and a' q w, the angles at p and v, at \ and q, and at 
B and w are respectively equal. Hence the triangles q w a and 
A B p are similar, and it follows that the angle q a p keeps constant 
in any motion of the mechanism, and the ratio of the lengths a q 



5 so APPLIED MECHANICS. 

and A p keeps constant. It follows, therefore, that if a is fixed and 
r is allowed to follow any cnrved path, q will follo^i^ a similar path. 

472. Note that Fig. 334, where p and o. are in the straight lines 
connecting w v and b v, is only a particular case of Fig. 336. Note 
also that if a w and w v are a pair of links guided at a and v along 
any paths, and if a b and k y are another pair of links whose A and 
Y follow identical paths with the first pair's A and v, then for each 
point in the one mechanism there will he a corresponding point in 
the other which follows a similar path. 

It is interesting to prove that if the links a w, w y, a b, b v, 
forming a parallelogram as in Fig. 336, he joined tip as in Fig. 337, 

then any f oiu' points p q r s in a 

B . w line parallel to av remain in a 

p//\^^><^^^/\\_ line parallel to av, however the 

y^^^-^^ ^^^^^\\. mechanism may alter in shape ; 

y:!^^^ ' • ^^■^^^^x and the distances p q, q r, r s are 
^-- — " y always such that p r . p q is con- 
Fig. 337. stant. Also pa = r s. 

473. We study mechanisms, 1st, 
Because in desig-ning new machinery we ought to have a general 
fairly exact knowledge of the sort of motion which each piece has 
in existing machines. For this we watch existing machinery, and 
work with rude models whose dimensions may he varied. 2nd, 
Because we wish to perfect an existing mechanism. For this we 
need a more exact study of the motion of every part, and we use 
skeleton di'awings. algebraic and trigonometric analysis, or graphical 
methods such as I have described. 3rd, Because in these days of 
increasing speed of machinery, the forces necessary to produce 
the accelerations of all the parts have become important, both 
for the strength and stiffness of all the parts, and also for the 
effects of A'ibration. 

474. In Fig. 329 we saw that, knowing the directions of motion of 
w and V, we can find the instantaneous centre. Now let w v get any 
motion whatsoever in the plane of the paper. Let the velocities of 
w and A- be %v and v in the directions marked. If we draw perpen- 
diculars to %v and f, meeting at c, the angular velocity of w about 



Fig. 338. 

c must be the same as that of v, and this must be the same as that 
of any other point k, rig-idly attached to w, about c. Hence, join 
any point k to c, the velocity of iv is at right angles to c k, and 
is represented in amount by the length of c k. Hence, as soon as 
we find the instantaneous centre, we have a diagram of velocities 
of all points in the body. Notice that if we have the instantaneous 



APPLIED MECHANICS. 



581 



centre c and the velocity of any one point, we have tne whole 
diagram to scale. If c d is a perpendicular to v w, as the Icng-th 
of* every line now in the diagram represents the amount of a 
velocity which is at right angles to its direction, c d is the com- 
ponent or common velocity of all points in the straight line w v in 
the direction w v. d w is the velocity at right angles to the direc- 
tion w V of the point w, and 

D V is the same for v, hence !^ ^^^ 

"W V represents v's velocity in 
excess of w's in this direc- 
tion. Hence, if w v, v x, x y 
are given links, and if we 
know the directions of the '^~ 
velocities of each joint at the Fig. 339. 

instant, say the directions of 

the dotted lines, we can find the velocity of every point in any 
body which is rigidly a part of any of the links if we know the 
actual speed of any one point. 

Choose a pole o. Draw lines fi^om o at right angles to the 
velocities of all the joints — o iv, o v, etc. Let the distance along 
any one of them represent the velocity of that joint to scale, and 
now draw the lines wv, vx, xij parallel to the real links. Then the 

lengths o w;, o v, o a;, oy re- 
Jk ^ present the velocities of the 

joints. To prove this, drop a 
perpendicular o d from o ujion 
any of the link directions in 
our diagram ; we choose vw. 
If o t^ is the amount to scale 
and is at right angles to the 
velocity of w, then o d repre- 
sents the velocity of w in the 
direction of the link w A'" ; hut 
this ought to he the same for v. 
We have, then, the amount 
o D of a component of v's 
velocity, and we have the 
direction of its whole velocity, and our construction is the very 
one which we should adopt to obtain v's velocity. If w and y 
are fixed, the diagram wvxy is a closed polygon, in the present 
case a triangle. 

Notice that in any case, if from o we drop a perpendicular o d 
on a side of the diagram, say wv^ then d w and d v represent the 
velocities of w and v at right angles to the length of d w. Hence 
the diif erence wv, divided hy the actual length of w v, represents 
the angular velocity of wv. Thus, in Fig. 343, take awvb as 
four links, A and b having no velocities. From the point o, which 
we may also call a or &, draw lioes parallel to aw and b v, because 
we know that these are at right angles to the velocities of w and v, 
and draw vm parallel to v w. Then ^u -r b v, aw -^ a w, and 
vm -T- V w represent the angular velocities of the three links. 

475. In Fig-. 339, if k ifl a point rigidly attached to the link vx 




Pig. 340. 



582 



APPLIED MECHANICS. 



(in the diagram find i& so that rkxis a triangle similar to t k x}. then 
ok is at right angles to and represents to scale the relocitv of k. 
Professor K. IB.. Smith, who vas, I teliere, the inrentor of thia 




Fig. 342. 

metiliod, takes his radiating lines pais^Del to, instead of at right 
smgies to. flie actual veloeities (Tr^ia. BjS.E., Jan., 183-5). He 

^1 T^ jinilir ::r.r'T:i:"i::i f rr^erations. 

Ir 1 — ~ ' '-' -~-- ly rotates in the plane of the 

- - - '-' 1" T - '^7:>ints A and k: are in the 

:>rtional to the distances 

- —iih a motion in the plane 

of:- 1 1 : : 1 eelerations of A and 

bct — - It^ .: - - u: lirectioai by the lines 

: :.:. 1 IV : ': the fignre aJtb 

exactly :_: : the real body 

A KB. Thi u^i :. lcz:.:i n of any point 

X is represented in amonnt and directian 

by o i. 

To prove this : The acceleration of a is 
the Tector sum. acceleration of b -|- accelera- 
tion of A relatiTeiy to b. Hence ba repre- 
sents this acceleration of a relatiTeiy to b. 
Bat the motion of a and Uie ^diole body 
relatively to b is a rotation, and heace hi 
represents the acceleration of k relatively 
to b to the same scale. The vector snm 
ob -^ bk = ok r^resents tberefore k*8 ao- 
<^ \ celeration. 

Notice that the true acceleration of any 
point a is the vector snm of two accelera- 
tions — ^the first in the direction of motion. 
Fig. 343. the second the centripetal acceleration if 

the path of a is cnrved. 
476. Let a known force in any direction be applied to a, a point 
in the body a k b which has a known motion in the plane of the 
paper. 

1st- Imagine the body divided into small equal masses. The 
construction of Fig. 342 enables us to find the forces with which 
these masses resist their accelerations. They may be combined 




APPLIED MECHANICS. 583 

with the weights of the parts to get what may be called the load 
diagram of the body. 

Let it be known that there is another force of unknown amount, 
but known in direction, acting at a (for example, the guiding force, 
with or without friction, at the crosshead of a steam engine). It is 
easy to see that, by Art, 475, we can find the amount of this 
guiding force at a, and also the total force which must be acting at 
a given point b. 

2nd. In Ai^t. 475 we have a construction which enables us to 
find the stress at any point of any section of the structure a k b, if 
it may be imagined to be any quasi-prismatic structure like a beam 
or connecting-rod, or even if it be shaped like part of a metal arch. 
In such a case as that of the connecting-rod, the steam-engine 




Fig. 344. 

maker ought to study the motion from many points of %'iew, o v 
and vw iDeing the crank and connecting-rod in any position. 
Produce w v to meet o b in b. o b is at right angles to the line of 
centres, o w. Then the lines o v and o b are at right angles to the 
velocities of v and w, and v b is in the direction of the link w T ; 
consequently, ovb is (Art. 475) a diagram of velocities, ov 
representing to scale the constant velocity of v, o b represents to 
the same scale the velocity of w ; also the distance vb represents 
the angular velocity of the connecting-rod. It follows from this 
at once that a force f at w in the direction w o would, if there 
were no friction and the connecting-rod were massless, produce a 
turning moment f x o b on the crank shaft. 

It can be shown that if we draw b a at right angles to w b, join 
V A, draw b c parallel to v a, let tv be in w o as far from a as c is, but 
on the opposite side of A, and join w v, then o w v is an acceleration 
diagram such that if k is any point in the connecting-rod, and we 
let the points w, k, and v be relatively to each other as w, k, and v, 
{k is evidently in the same horizontal as k), then k o represents the 
acceleration of k in direction and magnitude to the same scale to 
which V o represents the centripetal acceleration of v. In fact, 
o IV ky is a diagram of accelerations. 

477. When one point of a body is fixed, and we may neglect 
centripetal accelerations, we may take up the problem as a particular 
case of the general problem akeady considered, or we may attack 
it analytically as follows : — If o p is the straight centre line of an 
arm, o being fixed, and if a is its angular acceleration ; neglecting 
forces ]«arallel to o r, let there be a force r acting at r ; the 
acceleration at any point q (neglecting radial acceleration) is xa, i*f 
Q is X. If m is the mass per imit length, the load due to 



584 APPLIED MECHANICS. 

acceleration per unit leng-th is max; and if m is the bending 
moment at the cross- section (see Art. 357), 

^ = "'«-^- (1). 

Assiime m to "be some function of x, and let s be the shearing 

force at a section. Let I mx . dx = v, \ v . dx = «, and let the 



I mx . dx = V, 1 



values of v and u become vi and //,, when x = I. Then 

' = -^ ^ "'■ + ^ • • • ■ (-)' 
u = nil + ex + e . . . . {Z), 

F = , ?Jj + C . •. C = F - a Tj, 

D = ai(i + iv - i v\) I + c .'. e = a [hi - Ui) - F e. 

If A is the area of cross-section and p is the mass per unit volume, 
Ap = in. If we wish to have the same maximum stress /in every 
cross-section, and if z is the strength modulus of the section, then 
M = z/, 

m A p 
Dividing, therefore, (3) by m, we know the value of z/a every- 
where, if the arm is to be of tmiform strength. Thus, for example, 
let the section be rectangular, of breadth c and depth (in the plane 
of motion) ij. Then a = cy, z = J c:^^ ; so that z/a is ^y. Hence 

n ,,, T /)/ M 6 p M 

irom (4) we have — = - or y = — T -, or 

y = j^ {o.n + ex + e) .... (5). 

Since pzii = m, z = — , so that "uhen we know ?/ we know z : 

. = /,'^-....(e). 

Exainple.—LtQi m = a - bx, 

V = 1 ax^ - - 1 hx^, vi = I rt^ - 1 hl% 
u = lax^ ~ ^ bx\ ni = I al^ - ^ bl^, 
e =F - al^a - ^bl), 
e = aP {^ a - ^ bl) - T I, 

{an + ex + e). 



^ f[a- bx) 

478. In discussing the reciprocating motion of a j^oint, I have 
found in practice much simplification in my ideas when I have 
reduced the motion to some such shape as 

X = a sin. [qt + e{) + b sin. (2 qt + e^) + etc. 

where x is the displacement of the point from some fixed point in 
its path, and </ is 2 tt multiplied by the fi-equency, or 2 tt divided by 
the peiiodic time. For many purposes we find the first term 
sufficient. In most valve motions, such as link motions and radial 




APPLIED MECHANICS. &85 

gear, I find that'only two terms are ever needed even for a very 
exact definition of the motion. One interest attaching to this 
method of working is in its showing how the b term becomes 
twice as imj)ortant in the velocity, and four 
times as important in the acceleration as the 
fimdamental term. Another important matter 
is this : in modern machinery vibration is 
becoming very important, and the above de- 
scription of a motion is the one that lends 
itself most easily to a discussion of the vibra- 
tions which want of balance gives rise to. (See 
the author's books on the Calculus and on the 
Steam Engine.) 

479. The following proposition is the 
foundation of most calculations on motion 
communicated through links. The points a c b ^ 
are ia a straight Hne, They move, keen- Fig- 345. 
ing in a straight line, and at the same dis- 
tances from one another. (In engineer's language a c b is a 
moving link, and c is a point in it.) Prove that the motion of c 

Ac 
is the vector sum of the fraction — of b's motion, and the fi^action 

AB ' 

- — of a's motion. For let A a', b b', c c' be any displacements of a, b, 

and c. Join a' b. Draw c c" parallel to a a', and join c" c'. Now 

— = -7 — = —. — ;, so that c" c' is parallel to b b'. Hence, as c c 
a B A b A B ^ 

is the vector sum of c c" and c" c', we have proved the proposition. 
It is easy to extend our reasoning to motion which is not 
parallel to one plane. The motions of any three points a b c of the 
body, not in one line, define the whole motion ; and when we are 
given the accelerations of a, b, and c, it is easy for anyone who 
knows descriptive geometry to make a diagram showing the accelera- 
tion of any point of the body. 

480. Newton's great one law of motion for any, however complex, 
system of bodies is this : Look upon the rate of change of 
momentum of any small portion of matter of a system as a force 
in the opposite direction. All such forces are in equilibrium with 
the forces which act on the system from the outside. In most 
cases it is this most general way of stating the law that is most 
useful to engineers. Given their diagram of accelerations, they 
really have a force or load diagram, and the problems to be dealt 
with are now merely worked out by graphical statics. 

481. Most men are led to their study of this subject through what 
is called D'Alembert's principle. This is a principle which served 
a very useful purpose. At a time when English mathematicians 
were stagnating, being academically learned as to Newton's 
methods, but being really ignorant of them, the French mathema- 
ticians were developing kinetics practically independently of 
Newton's methods. Newton's third law was quite misunderstood, 



586 APPLIED MECHAXICS. 

and D'Alembert discovered a principle wkicli gave the power ol 
sohing dynamical problems with, certainty. It will not astonish 
am-one who knows academic methods to he informed that, although 
D'Alembei-t gave what is now seen to he merely a rather cumbrous 
explanation (easily misimderstood, because of certain technical 
tenns which may be confounded with one another) of Newton's 
third law, it is through the clumsy explanation that the subject is 
nearly always approached. We believe that this is one of the 
greatest reasons why many engineers are so disgusted with higher 
studies in dynamics. There seems to be as much absence of 
common sense now in academic persons as there was in the time 
of Erasmiis. He, the greatest scholar of the fifteenth century, 
wrote : " They are a proud, susceptible race. They will smother 
me under six hundred dogmas. They will call me heretic, and 
bring thunderbolts out of their arsenals, where they keep whole 
magazines of them for their enemies. Still, they are Folly's 
servants, though they disown their mistress. They live in the 
third heaven, adoring their own persons, and disdaining the poor 
crawlers upon earth. They are surrounded with a bodyguard of 
definitions, conclusions, corollaries, propositions explicit, and 
propositions implicit," 

482. Let the engineer take Newton's law in its very simplest 
form, as above expressed, and he will have no dilhculty in attacking 
the most complicated problems, for the dynamical becomes a static 
problem on the equilibrium of forces. It is usual to express part 
of the result analytically in the following way : — If in any direc- 
tion, which we may call x, the small mass m has the acceleration x, 
then the resultant force acting from outside the system in that 
direction is equal to the sum of all such terms as mx. State this 
as being true in any three directions, and of course it is true in 
any direction whatsoever. 

Now if the X of the centre of gravity of the whole system is x, 
we know that x'Sm — ^mx. Differentiate with regard to time 
once, and again, and we see that the whole mass 2>/*, multiplied 
by the acceleration of the centre of gravity, is equal to the sum of 
all the masses multiplied by their accelerations in the direction x. 
It follows, therefore, that the motion of the centre of gravity of 
a system is the same as if the mass of the system were collected 
there, and all the forces acting from outside on the system 
acted there. The other part of Xewtun's kiw is, of coui-se : 
The resultant moment of all the outside forces about any axis 
is equal to the rate of change of moment of momentum of the 
whole system about the same axis. We very often choose as our 
axis an axis through the centre of giuvity, but it is well to notice 
that this is not necessary. 

483. Angular Motion. — We know (Art. 92) that when a rigid 
body can only rotate about an axis, if the sum of the moments of the 
forces acting on the body is m, when the body moves through the 
angle 50 radians the work done is m . S0. If any little portion m of 
the msv^s of the body is at the distance r from the axis, and the 



Af>PLIED MECHANICS. 587 

body is rotating- with the angular velocity «, so that ar is the 
velocity at the place, the energy stored up in the little mass is 
\md^r^. If this is summed up for the whole hody, we see that 
every little term contains \c^\ so that if we know the simi of 
all such terms as mr'^ (called i, the moment of inertia of the hody 
about the axis), we can say that the kinetic energy is Jia^, When 
a force f acts on a body through the distance 55, the work done 
being f . 5s, and the kinetic energy of the body being \ mv^, by 
assuming that work done is equal to gain of kinetic energy, we 
are led to the law -p = in x acceleration. We have an exactly 
analogous set of terms for angular motion. The work m . 50 
corresponds with p . S.s ; the energy \ laF' corresponds with J mv^. 
The proof, therefore, is exactly as in Art. 497 — namely, if a moment 
M acts, through the angle 50, on a body moving with the angular 
velocity a, and whose moment of inertia is i, so that its kinetic 
energy is \ icfi, and if a + 5a is its new angular velocity, then 

M . 50 = Ji (a + 5«)2 - ^laP' = 1 . ala + ^1 . {^af, 
Za , 5a - 
50 ^ 50 
As 50, and therefore 5«, become smaller and smaller, we have 
ha 50 5« 
^ = ^^-50 = 's^-50' 
or 

da 

" = ■<;« = " 

if a is used for — , or the angular acceleration. This is exactly 

analogous with the force law in linear motion, i « is called the 
moment of momentum of the body, and i a is the rate of change of 
the moment of momentum. 

484. To arrive directly at the moment of momentum of a rigid 
body about any axis, consider a portion of mass m at the distance r 
from the axis ; r making an angle with a iixed plane thi'ough the 

*^^' X = r cos. e, tj = r sin. 0, 

dx . „dd dy ^ de ,^. 

— = - r sm. — , ;; = r cos. . — . . . . (1). 
di dt dt dt ^ ' 

The moment of the momentum m -r about the axis \s> - my . -j-vx 

dt dt 

the direction of increasing 0, and the -moment of momentum of 

w ^ is mx . ,f , and the sum. of these is 
at at 

and the sum of all such terms is i-^-, or i a. The rate of change of 

at 

this is I -j-g , or i a. This also may be obtained by differentiatuig 
at 



APPLIED MECHANICS. 

dP'X di^v 

(1), and so finding -j-^ and ~. Taking tlie moments of the 

virtual forces - m-^-^ and - m-^i about the axis, we arrive at 

d-d 
- I— 2, or - la, whicli, together with the moments due to the 

outside forces, produce equilibrimn; or, with signs changed, are 
equal to them, as above. Students ought to think of other ways 
of reaching these results concerning rigid bodies. 

485. Kinetic Energy of any System. — Let m be the mass of a por- 
tion of the system in the position a;, t/, z. Let x, y, z at this instant 
be the position of the centre of gravity. The whole kinetic energy 
is the sum of such terms as 



m-m-mv 



Let X = X + x^, y = y + 'ip-, z = z + z^ \ s,o that x^, y^, z^ show 
the position of in relatively to the centre of gravity. Note that 

/dx\^_/dx ^^_{^_^\^ o^f ^^^ /dx^Vi 
\dt ) ~ \dt "^ dt ) ~ \di) '^ ^dt'U + \dt) ' 

But 'Zm . 2 — . — =2 — ^m . — , and 2 m — is because it i& 

dt dt dt dt dt 

the rate of change of 2 m x^ and '%mx^ = 0. Hence 

2: 






The first of these terms is the whole mass multiplied by the square 
of the X component of the velocity of the centre of gravity. It 
follows, therefore, that the kinetic energy of any system is 
equal to the kinetic energy which the system would have if it were 
all moving with the velocitj' of the centre of gravity + the kinetic 
energy due to the motions of all the parts relatively to the centre 
of gravit5^ 

Now, the motion of a rigid body relatively to any point of it 
can only be a rotation. Hence for rigid bodies we have the rule : 
If the velocity of the centre of gravity is v, and if there is a 
rotation of angular velocity Q about some axis, and if Iq is the 
moment of inertia of the body about a parallel axis through the 
centre of gravity, and m is the whole mass, the kinetic energy e is 

\ M V- + I Iq (0)2. If lo = m7^2^ then e = ^ m | v^ + 7,-2 (0)2 | _ it 

is only in the case of a rigid body (in all this by rigid body I 
mean an infinitely rigid body) that we have this simple rule. In 
the case also of a rigid body we can find law (Art. 482) from the 
law of energy, as we do in Art. 495 ; but we cannot do this so 
easily for a system having internal relative motion, because there 
is internal potential energy, which may alter. The virial and 
other laws, which may easily be arrived at, do not concei'n 
engineering applications of mechanics. 

\ 



APPLIED MECHANICS. 



589 



486. In any ordinary elastic body the internal motions are vibra- 
tional. The momentum and moment of momentum of the system 
are therefore practically the same as if the body were quite rigid. 
This is not the same in regard to the energy. We say that it 
disappears or is changed to heat and other forms ; this means 
merely that it has become molecular, and equations which regard 
the body as if it were rigid are quite inapplicable. Thus it is that 
the momentum equations are what we rely upon, because when we 
study the motions of bodies momentum cannot be hidden as energy 
may be. 

487. "V\Tien we work exercises on rigid dynamics we ought to 
apply Newton's law always in the shape given in Art. 480. When 
we work a new exercise let it be regarded as a new illustration, to be 
worked to making our knowledge of the fundamental principle 
clearer. Most details of one's theory and artifices for the solution of 
problems will disappear from view in one's professional work. Let 
the fundamental notion be so well fixed that it cannot disappear. 

488. Exercise. — In the compound pendulum of Fig. 346 find the force act- 
ing at s, the point of suspension. In Fig. 347 let g be the centre of gravity. 




Fis. 346. 




Let X and y, as shown, be the horizontal and vertical components of the 
force at s. Let g s y be 0, and let the body be mo^dng so that d increases. 
If s G be called r, the velocity of g in its path is r9. It has an accelera- 
tion rt/ in its path, and ^(O)^, a centripetal acceleration in the direction 
G s. L Kegard now the whole system of forces, which are supposed to 
balance if they were to act at one point, g. Let m be the mass or w/^, 
we have - m rd in the direction g a, - m ;-{0)2 centripetal in the direction 
G s, w downwards (the weight of the body), x horizontally, and y down- 
wards. Eesolve these in any direction whatsoever, they are to balance. 
Thus, resolve them horizontally and vertically, 

X - la.rQ C0S.9 + m r{9f sin.9 = .... (1), 

Y + w + Mr0 sin.0 + ur{ef Gos.9 = . . . . (2). 

Observe that the forces in g a and g s are virtual forces, or forces 
equal and opposite to mass multiplied by acceleration, II. Now take 
moments about any axis. The most convenient seems to be s. w is the 
only externally applied force that has any moment about s, and so 

w *• sin.0 + I'e = . . . . (3), 



590 APPLIED MECHANICS. 

if I is the moment of inertia about s. Frc^m these equations x and y, in 
terms of 9 and 9, may be calculated. We fii-st express 9 in terms of d 
from (3). If I = - k-, (C) gives 



>in. 



9 

Y = - W 



,.0f- p" sm.9\ - - r sin.9 . [Bf, 

^rsm.ef - ^i; sm.e) - - ;• {df cos.9, 
y \ k- / ff 

= - ^, sin. 20 - -sin.0(0)2, 



1= - n-!i!i^J-^-^ cos 0-0)2. 

9 ■^'ill depend on the limit of the swing, and mav have any value. 

A 



^i o 



Fig. 348. 

489. Vibration Indicator. — Fig. 348 shovrs an instrument which 
has been used for rudicarmg quick vei-tical vibration of the ground. 
The mass c p q is supported at p by a knife-edge, or by friction- 
wheels. The centre of gravity g is in a horizontal line with p and 
Q. Let T G = a, G Q. = b, T Q. = a + b = I. The vertical spring 
A E and thread r q support the body at q. As a matter of fact, a b. 
is an Ayrton-Perry spring, which shows by the rotation of the 
pointer r the relative motion of a and q. Let us neglect its 
inertia now, and consider that the pointer faithfully records 
relative motion of a and q. It would shorten the work to only 
consider the forces at p and q in excess of what they are when in 
equilibrium ; but, for clearness, we shall take the total forces. 

^lien a body gets motion in any direction parallel to the plane 
of the paper, we get one equation by stating that the resultant 
force is equal fnumerically) to the mass multiplied by the linear 
acceleration of the_ centre of gravity in the direction of the 
resultant force, "^^'e get another equation by stating that th* 
resultant moment of force about an axis at right angles to the 
paper through the centre of gravity is equal to the angular 



APPLIED MECHANICS. 591 

acceleration, multiplied by moment of inertia about this axis 
through the centre of gravity. I shall use x, x, and x to mean 

displacement, velocity, and acceleration, or x, — , and -to- 

do Cto 

Let p and a have a displacement Xj^ downward. Let q be 
displaced x downward. Let the pull in the spring- he q = Qq + c 
{x - Xi) where c is a known constant {c is the reciprocal of the h 
used in Art. 514). Let av he the weight of the body. Then if p^ 
and Qq ^6 the upward forces at the noints marked p and Q, in the 
position of equilibrium, 

% (» + *) = w a and Pq + q^ = w. 
Hence 

i w a^Y ... 

^0 = ^T^j. ^0 = ^Tjn • • • • (1)> 

01= Qo + e {x - xi). 

Now G is displaced downwards — -^^i + tX, so that 

■^ a + b ^ a + b 

The body has an angular displacement 9 clockwise about its centre 

of mass, of the amount p. So that if i is its moment of 

' a 4- b 

inertia about g, 

. ^h + ^a = j^^(x - X,) . . . . {Z), 

Hence (2) and (3) give us, if m stands for — , and if i = m F where 
k is the radius of gyiation about g, 

If ^1 is the radius of gyration about p, we iind that (4) 
simplifies to 

X 4- ^^^ = e^Xi + n^Xi .... (5) 
if n stands for y- / — = 2 tt x natm-al frequency and e^ stands 
for 1 - 272- ^^ X - Xihj the letter y, because it is really ^ that 



J 



592 APPLIED MECHANICS. 

an observer will note, if the framework and room and observer 
have the motion Xu Then, as t/ =^ x - Xi, ot x = i/ + xiy 



1/ + Xi + n- (y + z^) = c^xx + nH\. 



So that 



y a. n-y = {e^ -\)xx.... (6), 

or 

f/+«-i/ + ^72^1 = 0....(7). 
Thus let x-^ = A sin. qt. 

We are neglecting friction tor ease in imderstandtng onr 
results, and yet we are assuming that there is enough friction to 
destroy the natural vibration of the body. AVe find that if we 
assume i/ = a sin. qt, then 

_ al q^ 
«- U^-n^ - q^^' 

That is, the apparent motion y (and this is what the pointer of an 
Ayrton-Peny .spring will show ; or a light min-or may be used to 

thi'ow a spot of light upon a screen) is ^^ -^ — ^ times the actual 

potion of the framework and room and observer. If ^^ is large 
compared with n — for example, if ^ is always more than five times 

ix — we mav take it that the apparent motion is ^r, times the reaJ 

motion, and is independent of frequency. Hence any periodic 
motion whatever (whose periodic time is less than 4th of the 
periodic time of the apparatus) will be faithfully indicated. 

Xote that if ff/ = h-^, so that q is what is called the point of 
percussion, q is a motionless or "steady" point. But in practice 
the instrument is very much like what is shown in the figure, and 
Q is by no means a steady point. Apparatus of the same kind may 
be used for east and west, and also for north and south motions. 

490. We did not think it necessary to interrupt our accoimt of 
simple harmonic motion to speak about the analogies between 
linear and angular motion, although we began to use them in Art. 
•453. We see (Ajt. 482) that any motion whatsoever of a rigid bcdj 
may be most simply studied as a motion of translation and a 
motion of rotation, and we shall find it best to take the translational 
motion of the whole body as the actual motion of its centre of 
gTa^ity, and the rotation as one about an axis passing through the 
centre of gra-^dty. When we do this, we shall find the following 
angular formulge valuable, not merely for the motion of a body 
whose axis is fixed, but for any motion whatsoever. The only 
point in which the analogy between linear and angidar motion, 
fails is this : The mass or inertia of a body is independent of the 
direction of motion ; the moment of inertia of a body is usually 
different about different axes of rotation. 



APPLIED MECHANICS. 



593 



COMPARISON OF LINEAR AND ANGULAR MOTIONS. 



s = space, t = time. 
ds 



V — velocity 



de 



a = acceleration = -t3, or s, or v. 
dfi' ' 



m = mass or inertia = tv/ff. 

F = force. 

mv = momentum. 



If a body has two disi)lacements 
(or velocities, or accelerations, or 
momenta) given to it simultane- 
ously, represented in magnitude 
and direction and sense by o p and 
o ft, they are equivalent to the 
displacement (or velocity, or ac- 
celeration, or momentum) o r if o p 
and o ft are the sides, and o r the 
diagonal of a parallelogTam, the 
arrows being all difluent or con- 
fluent. 



Forces represented by op and 
o a have a resultant o r. 




a = F — m. 

Impulse = change of momentum 



=J 



F . dL 



Average force = change of mo- 
mentum -~ time. 

Work done = force x distance. 

Space average of force = work 
done -f- distance. 

Kinetic energy, | mv'^. 

Under uniform acceleration 
from rest at time 0, v = at, s = 
I at% v^' = 2 at. 



angular space. 



do 



a = angular velocity = -3-, or d 



dV 



a = angular acceleration 



d^'d 



or 



e, or a. 
I = moment of inertia. 
M = moment of forces or torque. 
la = angular momentum, usually 

called moment of momentum. 

If a body has two angular dis- 
placements (or velocities, or accele- 
rations, or momenta [moments of 
momentum]) about the axes o p and 
o ft, and if the lengths of the lines 
o p and o ft represent these to scale, 
and if the arrows indicate positive 
direction, as that in which a right- 
handed screw moves, they are 
equivalent to the angular displace- 
ment (or velocity, or acceleration, 
or moment of momentum) repre- 
sented in axis, amount and sense, 
by OR. 

Moments of force or torque 
represented by p and o ft (the 
direction of the line being the axis 
about which the moments of forces 
are taken, and the sense of the 
arrow-heads, as in the last case) 
have a resultant o b, 



a = M -^ I. 

Angular impulse -— change of 



moment of momentum 



I" 



dt. 



Average torque = change of 
angular momentum ~- time. 

Work done = torque x angle. 

Angular average of torque = 
work done —■ angle. 

Kinetic energy, 1 1 a^. 

a ~. a', e = laf^, a? = 2 at. 



594 



APPLIED MECHANICS. 



Simple harmonic motion if t is the periodic time, a is the amplitude, 
e is the lead 



r = A sm. 






27r 

or a = A — cos 



__ s or _ t2 

A body moves backwards and 
forwards imder the action of a 
variable force, which is always 
proportional to the distance of the 
body from its middle position, and 
which always acts towards this 
position; and if the force at a 
distance of 1 foot is 5 lbs., then the 
time of . -vibration is 2 tt times the 
square root of the quotient of the 
mass of the body divided by 5. 



If m is the mass ; if the force of 
friction is h times the velocity; if 
the constraining- force is n times 
distance from centre, 



If a body vibrates about a fixed 
axis under the action of the torque 
(say from a sjDiral spring or twisted 
wire), so that it is always propor- 
tional to 0, the angular displace- 
ment of the body from its mean 
position, and if the torque is 5 
pound-feet when the body is 1 
radian from the mean position, 
then the time of a vibration is lit 
times the square root of the quotient 
of the moment of inertia divided 
by 5. 

If I is moment of inertia, fric- 
tional torque is h times the angular 
velocity. If the constraining torque 
is n times the angular distance 
from the mean position, 



s ox Q = Kf:~^^ sia. [dt + e) 
where/ = \ him, or ^ bji, and 



2^V4>/— b'^^ox2-^j4.n'^-b\ 



A body of mass m moves ^vdth 
constant speed v. At any instant 
its momentum is in the direction 
OK, and is represented to scale by 
the amount o e. If a centripetal 
force F of constant amount is always 
acting at right angles to the direc- 
tion o E, the effect is to make the 
point E travel with linear velocity 
equal to r. That is, the direction 
of motion alters with an angular 
velocity f/«2V. 



A body whose moment of mo- 
mentum is represented to scale, 
and the direction of its axis by o e, 
is acted on by a constant torque m, 
which is always about an axis at 
right angles to o e. The effect is to 
make the point e travel with linear 
velocity equal to m ; that is, the 
axis OE has an angular velocity 
(towards the axis of the torque) 
whose amount is m -j- moment of 
momentum la. 



APPLIED MECHANICS. 



595 



EXERCISES. 

1. A top (Fig. 351) rotates at r radians per second about its axis o c. 
The axis rotates (or precesses) at n radians per second in a conical path 
round the vertical o z. Represent this motion by one cone rolling on 
another. 



— -E' 



•2 H 




For M 
O 
Fig. 350. 

Answer. — If o c is the axis of the rolling cone of angular velocity r 
of which o I is the slant side, if o z is the axis of a fixed cone of which o i 
is the slant side, at any instant o i is a line in the rolling cone, which is at 
rest, and about which it rotates with angular velocity w. Let z o c = a, 
I o = j8. Draw j k and j l at right angles to o z and o i. We see 
that J would go into the paper w^th the velocity n . k j, because j is 
moving in a horizontal circle whose fixed centre is k ; also j goes into 
the paper with the velocity co . l j, because l is fixed ; and these are equal ; 
so that n . K J = a> . L J, or, since k j = o j . sin. a, l j = o j . sin. /3, 

n sin. a = o) sin. fi . . . . (1). 

Again, a> about o i is the total spin of the top. It has components 

w cos. fi = r about the axis o c ) .^. 

w sin. ^ or r tan. j8 about the axis o a j ' ' ' ' ^ ''' 

if o A is at right angles to o c ; and hence (1) may be written 

n sin. a = r tan. ^ . . . . (1), 

io that /8 is now known. All the lines drawn are in the plane of the 
paper at the instant. 

2. If the moment of inertia of the top about o c is c and about o a is a, 
find the resultant moment of momentum. 

Answer. — c . r is the moment of momentum about o c ; a . r tan. ^ is 
the moment of momentum about o a. The resultant has an axis o h, its 
amount being c r . sec. y. If y is h o c, 

tan. y = Ar tan. ^ ~ cr = a tan. yS -J- c . . . . (3). 

3. If the distance oh represents the moment -of momentum of a top 
about the axis o h, and a torque ^s' h sin. a about an axis at right angles 
to the paper tends to send o h away from o z, the effect is to make the 
horizonijfcal velocity of the point h at right angles to the paper be w A sin. a. 
What is the angular velocity of o h about o z ? 



596 APPLIED MECHANICS. 

Answer. — ^The point h moves in a circle of radius o h . sin. z o h, ana 
the linear velocity of h, or w A sin, a di'^ided \yx radius gives ano[iilar 
velocity, or 

w h sin. a -7- o H sin. z o h, 

or n = ^^^^^"-^ or ^h.^m .a 

o H . sin (a — 7) cr . sec. 7 . sin. (a - 7) 

Hence Her (sin. o - cos. o . tan. 7) = w A sin. o. 

But tan. 7 we found to be a tan. )3 -7- c, and hence 

Clcr (srm. a - cos. a . - tan. j8 j = W A sin a. 

But (1) gives us tan. /3 = H sin. a'/r, and hence 

£iCT - n^ . cos. a . A = ^ h, n--n. — sec. a . + — sec. a = 

A A 

From this we may calculate Cl, the rate of precession of the top. The 
student will find a value of r for which any precessional motion is 
impossible, and the top must fall. He may also consider the meaning of 
two different real values for D.. \^e have given this exercise be<:ause it 
gives a good drilling in the use of the above rules. Xote that a real top 
has not an infinitely sharp peg. It rolls and slides on the table, and W9 
shall not consider this much more complicated problem. 



597 



CHAPTER XXYII. 

CENTRIFUGAL FORCE. 

491. Centrifugal Force. — If a body is compelled to move 
in a curved path, it exerts a force directed outwards from the 
centre, and its amount in pounds is found by multiplying the 
mass of the body by the square of the velocity in feet • per 
second, and dividing by the radius of the curved path. We 
evidently get the same answer if we multiply the mass by the 
radius and by the square of the angular velocity in radians 
per second. Thus, a weight placed at the end of an arm like 
the arm of a wheel exerts a pull in the arm. If a body moves 
round an axis 20 times per minute in a circle whose radius is 
3 feet, you can determine the centrifugal force by first finding 
the velocity of the body and using the above rule, or you may . 
proceed as follows : — The weight of the body multiplied by 3 
multiplied by the square of 20 divided by 2,937 is the centri- 
fugal force. 

Suppose a wheel, whose total weight is 20 tons, or 44,800 
lbs., has its centre of gravity 0*4 foot away from the axis — 
that is, suppose the wheel to be eccentric — then if the wheel 
makes 50 revolutions per minute, the centrifugal force is 
44,800 X 0-4 X 2,500 -^ 2,937, or 15,253 lbs.— that is, 6-81 
tons. This force acts on the bearings of the shaft, always in 
tlie direction of the centre of gravity of the wheel. 

492. Anyone who wants to get clear ideas about centri- 
fugal force ought to make experiments of his own. Unfor- 
tunately, although there are many toys made to illustrate the 
effects of centrifugal force, we know of only one piece of 
apparatus which enables the laws to be systematically experi- 
mented upon. Fig. 352 is a drawing of it. A represents a 
little, flat, cast-iron box, like a narrow drum ; one drum-head, 
as it were, being replaced by a thin steel plate, so as to be 
strong and flexible, b is a glass tube which enters the box. 
Mercury fills the box and the tube to the level 5, and when c, 
the centre of the steel plate, is pulled or pushed, although we 
cannot see much yielding in o, we can observe the mercury 
rise and fall in the tube. There is a screw, D; entering the 
box at the back ; by means of this screw we can adjust the 



APPLIED MECHANICS. 599 

height of the mei'cury in the tube. The box is in the centre 
of a circular table, E, which can be whirled round at any speed 
we please, and the tube is exactly in the axis of rotation, so 
that the height of the mercury can be measured whatever be 
the speed. Fastened to the centre of the corrugated plate at 
C is a long brass rod, F, which is supported at J on the end of 
a little rocker, so that it can move backward and forward 
with less friction than if it were made to slide on a bearing. 
At any place along f we can clamp a weight, h, which we may 
alter as we please from 0-5 to 8 lbs. We can clamp it near 
the axis or one foot away, the radius of the circle described by 
its centre of gravity being measured by the scale in tenths 
and hundredths of feet marked on the rod. The experimenter, 
turning the handle which drives the pulley g, keeps his eye 
upon the mercury level, and is able to maintain a very great 
constancy of speed. He counts the revolutions of his hand, 
and one of his companions takes the time, so that no speed- 
counter or indicator is necessary. Now, the centrifugal force 
due to the rod and sliding weight causes c to be pulled out 
very slightly, and this causes the mercury to fall in the tube ', 
and it is easy with a fixed vertical scale alongside the tube to 
measure this rise and fall. We usually get a spring balance, 
or a cord, pulley, and weights, and before our experiments 
begin we pull the end of the rod f, noting the height of the 
mercury for a pull of 1 lb., 2 lbs., etc., and in this way we can 
aftervv^ards tell the value of our scale measurements. We also 
make a number of experiments when the sliding weight is 
removed from the rod F, to tell us the centrifugal force of 
everything else at different speeds, and this we subtract from 
our subsequent observations. We see, then, that we can 
measure the centrifugal force, in pounds, of various masses, 
from 0'5 to 8 lbs., moving at any speed in a circle whose 
radius may be as little as 1*5 inch and as much as 11-5 inches. 
With this instrument it is easy to test the law which is 
usually given, and without working with some such instru- 
ment we question if students are likely to have any but vague 
notions about centrifugal force. 

493. There is another method of experimenting which 
suggests itself, with apparatus which anyone may fix up for 
himself, but it does not give such a thorough understanding 
of the law to the person who experiments. In Fig. 353, a is 
a leaden ball at the end of a silk thread, p a, fastened at P, 



600 APPLIED MECHANICS. 

A is kept out from its natural position in the vertical by 
means of a horizontal thread in the direction a b. Xow, if 
we pass the horizontal thread ab over a pulley and hang a 
weight at its end, we find that the force 
acting in A B is to the weight of a as the 
distance ka is to the distance kp. The 
, body A 3S acted upon by three forces : its 

\ weight downwards in the direction a a. the 

\ horizontal force a b, and the pull in the string 

\ A P. The triangle of forces tells us that as 

\ a K P is a triangle whose sides a k and k p 

\ and p A are parallel to the three forces, then 

— -0 ^__ the horizontal force, acting in a B, is to the 

vertical force which is the weight of a as 

! ^ the distance ka is to the distance pk. 

Fig. 353. Suppose, for example, the weight of a to 

be i lbs., the height p k to be 8 feet, and 

the distance a k 1 foot, then a k is one-eighth of K p, and we 

are sure that the horizontal force needed just to keep a in 

this iX)sition is 5 lb., for it must be one eighth of the 

weight of A. 

Now, let such a ball as a, hung by a thread P a, go round 
and round in a circle. Measure, as accui-ately as you can, k a, 
the radius of the circle, and k p, the vertical height fi-om tho 
ball to the point of suspension. Also count how many revo- 
lutions the ball makes per minute. The centi-ifugal force is 
now doing what the horizontal string did before, and we 
know how much it is. In fact, the centrifugal force is 
obtained by multiplying the weight of the ball by ka, the 
radius of its circle, and dividing by the vertical height K P. 
You can test if the centrifugal force law is tine, therefore, by 
means of your measurements. 

Convei-sely, if the weight of the ball is w, if a k is r, if p a 

w '*•■■ 
is I, and p K is A, the centrifugal force is - ^ , and we see by 

^ ' r 
the triangle of forces that this must l^e equal to j w. ZSTow, if 

the time of one revolution is T, then v = 2 ir r -r T, and hence 



becomes t = 2 tt a / i .... (1) 



w V' r w 

g r ~ h "^ 9 

A ball going i-ound in this way is called a conical 



APPLIED MECHANICS. 601 

pendulum, and we have in (1) found its periodic time. Its 
periodic time is that of a simple pendulum (Art. 446) whose 
length is the same as the h of the conical pendulum. 

EXERCISES. 

494. 1. A pendulum bob weiglimg 20 lbs. moves with, the velocity of 5 
feet per second at the middle of its path. What force is tending- to make the 
pendulum longer, in addition to the weight of the bob ? The length of the 

20 25 
pendulum is 15 feet. Ans., _ — x — = 1*035 lbs. 

^ ' 32-2 15 

2. A locomotive weighing 35 tons travels at 50 miles an hour round a 
ciu-ve of 2,000 feet radius. What is the centrifugal force? 

Alls., 2-92 tons. 

3. The centre of a ball of 10 lbs. is at 4 feet from an axis, and revolves 
at 500 revolutions per minute. What is its centrifugal force ? Its 
centrifugal force is to be balanced by those of two balls attached to the 
shaft, each of 20 lbs., revolving in circles whose planes are 4 feet and 
1 foot from the, plane of the first circle. At what distance is each of 
them from the axis ? Ans., 3,400 ; 0-4 foot ; 2 feet nearly. 

4. The rim of a pulley has a mean radius of 20 inches ; its section is 
6 inches broad, and average thickness J inch. If it revolves at 200 
revolutions per minute, what is the centrifugal force per inch length of rim? 

Ans., 18-6 lbs. 

5. A vehicle describes a horizontal circle of 600 feet radius with a 
velocity of 40 miles j)er hour; find the direction of the resultant due to 
gravity and centrifugal force. If it is a railway of gauge 1 metre, how 
much higher ought the outer rail to be than the inner one ? 

Ans., 10°- 1 with vertical ; 6"9 inches. 

6. A skater describes a circle of 100 feet radius with a velocity of 
18 feet per second ; what is his inclination to the ice? Ans., 84°. 

495. 7. The law of energy states that a body of weight av and 
miss M falling from rest without friction along any path, at any instant 
being at the height h and having the velocity v, has the sum of its 
potential and kinetic energies constant, or 

w A + ^ M 1)2 = constant .... (1). 
Show from this that force is equal to mass multiplied by acceleration. 
If s is distance along the path measured in the direction of motion, 

differentiating, we have w — . + m r . — = 0. But - dhlds is sin. a 
as (Is 

if a is the inclination of the path to the horizontal, and v = dsjdt, 

dv 
ds 



dv 
and hence w sin. a = mv -^ . . . . (2| 



dv dv dt dv 

or, since v — = v ~ . ~ = — , 

ds dt ds dt 

w sm. a = II ^ . . . . (3). 
We see, therefore, that the resultant force (w sin. a) acting on the body 



^ 



602 APPLIED MECHANICS. 

is equal to tlie rate of cha n R^e pa* seoctnd of the mc-mentnia. If o is 90*, 
so that the hi-dy mils verticallT. then -^ is csiHeti g. and (-3) is merely 

M =: ^;> (4). 

Students ought to famDiaiise themselves "with this and other ways of shov- 
ing that the law of energy leads to the law : force = mass x ao-<:ieleratiC'ii. 

496. We think that Newton's method, the Thomson and Tait method, 
is very much to be preferred to any other method of starting in the study 
of mechanies. Erery engineer ought t-o have his t and t' (the elementary 
treatise), a well-thumbed hoot. Xevertheless, it is good to Ic-ok at our 
fundameDtal notions from the energy point of view. The older j^i'M'fs 
of statical principles were based on the idea of wort, as in our Chapter 
XT. Assuming that mechanical energy is not lost by conversion into 
heat or other forms and that a total store remains constant, we get good 
worting views of our sol^ect. Undeilyiiig our notions are our ^ecoJa- 
tions as to how matt^ p^<Hins aitiactif»8; Irat leaving out ideas of 
^ha^al ^76^ and sffaain eraexgy and coaafining our atteotion to tiie idea 
y^t the sum of the potential and kinetic oiei^es of a syet^n remains 
constanl^ we l»,Te a wucluug idea of great iffiefnlneBB. TJnfavtnnatd j 
nymy lossn. f»gi^ liiat mefiianifal enagy may be etm^'earted into heat^ 
and so tiiey make sndi. mi^afees as to eal^daJe tibe fcnree adiog on. a pile 
beJng driven as tibe space rate of eonriaaon of enesrgy. 

497. Let us take ^^mOter example. If a cnostant force f acta in ihe 
iirectiiMi of motian of a hody for a time S< Ihrono^ a ^ace Ss, increasing 
its T^odty from r to r -F Sr, &e woik doaie hy ihe force is f . 2s, »id 
this is spent in iniseaang tibte kine^ energy of the hody frcKm \ wufi to 
I m (f -f ccf. Hence 

car F = wr — + I »? — ^r (1). 

As we imagine Ss, and therefoi^e ^r, to be smaller and smaller wifiiont 
limit, the last term in (1) gets to be nearer and nearer 0, and r gets to he 

better and better represented by cs'Sf. Hence 

_ f-s Bv ^r 

ir =: ffi — . -— , or w . — , 

ct cs St 

; : : : : : r is e':3^ual to mass multiplied by acceleration, oi force is the rate of 
: '- " ■ ^- r :•! mMnentum. 

-z:i 5 . S : long as we deal with force in the direction of the motic^n of 

; re is no difficulty in showing that the law of wort or the 

_ :. of iriec-hari'i-al energy leads to the rule, " Force is rate of 

c'z: -Zi :: i:^:r.Tr.:v.r:-. "^hen-a point is moving in a curved path, we 

c. _ - : : 1; . :_ : :=.: of the force in the direction of motion is 

f .: :/:::_ _ : f momentum in the direction of motion. Bat 

-'. : .: __ : _ ;_■ i_ :_: at right angles to this r Here it is necessary 

-It _ : r.r -.ige of dlrection of motion, and not merely of 

speed, indi:.- Ti - -: lorce is acting on a body. It is a matter of 

Domm.on oI^t. -: : _ _at a centripetal force is nscessary to keep a hody 



APPLIED MECHANICS. 



603 




moving in a circle, and that the body exercises a centrifugal force ag-ainst 
those constraints which compel it to move circularly. When a curved 
surface gradually changes the direction of a moving hody, these two 
forces act at the point of contact,* To get clear ideas, consider the 
conical pendulum of Fig. 354. The hody p revolves in 
a circle at constant speed. To keep it at the constant 
distance a p = r from the centre a, we know that, if it 
were at rest, a force f must act. But it is in motion, 
and yet keeps out at the distance r. There is a centri- 
petal force whose amount is known to us, w tan, a, and 
it is evidently balanced by what we can only call 
a centrifugal force of this amount which is due to the 
motion. Observe that we here have a case of centri- 
petal force acting on a body, creating no increase of 
kinetic energ}', and creating no change of any kind of 
energy, for when p comes round to the same position 
again ever5i;hing is just as before, although the force has 
been acting for a whole round. Mathematically there is no great 
difficulty. We assume that we have proved that forces in directions at 
right angles to one another may be studied independently and may be 
combined as vectors. Consider circular motion in the plane of the paper. 
Constant speed v of p means a velocity v cos. Q in the direction q, a and 
- V sin. in the direction a c. The two accelerations are - v sin. Q . a 

and - V cos. 9 . a, a being — , the angular velocity, or -;. If m is the 

mass of the body, we know that there are two forces acting upon the 

body : - mv sin. . a in the direction 
a A, - 7nv sin. 9 . a in the direction 
Q c. Forces always compound accord- 
ing to the vector law, as may be 
proved from the law of work (see 
Art. 495), and hence we see that the 
force acting upon p in the direction 
p a is a centripetal force mv7v or m 

— , and the acceleration of p is a cen- 

tripetal acceleration v'^/r. We have 
seen (Art. 493) another way of arriving 
at the amount of the acceleration. 
In the very same way it is easy 
to show that when a small body 
Fig. 355. moves, not merely in a circle, but in 

* People may have notions of force that seem to be quite different from 
a metophysical point of view, but which are really the same mathematically. 
Thus one man iDuts it that there is no such thing as force ; we have only mass 
X acceleration. Another says — "Yes, we have a centripetal force ; say, acting 
on a body which is whirled round at the end of a string, but it is not right to 
speak of the equal and opposite force, sometimes called centrifugal force. " Now, 
if we only think of the body, it may be enough to speak only of centripetal 
force and centripetal acceleration ; but a string in tension is really acted on by 
equal and opposite forces. If one of these is exerted towards the centre, the 
other is exerted outwards by the body, and we call it a centrifugal force. 




604 APPLIED MECHANICS. 

any curved path, the force acting upon it has two components — one 
in the dii-ection of the path, equal to its mass, multiplied hy the rate 
of change of its mere speed; the other at right angles to the path, 
equal to its mass, multiplied by its linear velocity, multiplied by its 
angular velocity. 

499. It is only when we have looked at the subject from the energy 
point of view, and in many others in which it will strike a thought- 
ful student to make experiments, that the beauty of Newton's 
generalisation comes home to us and we see how all the results of 
observation, experiment, and speculation are given to us in his 
statements. In fact, we gradually get to loiow that, whether 
acceleration of a small body is along the path or at right angles to 
it, force i^ equal to the vector rate of change of momentum. Now, 
this is Newton's definition of force, and we may begin our study 
•of mechanics from this point of ^dew. But speculation of the 
above nature is very far from being useless. 

Taking up our subject from the easiest point of view, accelera- 
tion is the time rate of change of velocity, and force is the time 
rate of change of momentum. We first consider acceleration in 
the direction of motion, and then, if a body changes the direction 
of its velocity, the lateral rate of change of velocity or lateral 
acceleration must be considered. A velocity is a vector quantity, 
and velocities are added as all vectors are added. A velocity of 
5 feet per second eastward, added to a velocity of 5 feet per second 
southward, are equal to a velocity of 7 "070 feet per second south- 
eastward. 

500. When I speak of the motion of a body, I usually mr-an the 
motion of its centre of mass. If a body is at p moving along a 
curved path with the cor stant speed of v feet per second, when it is at u 




Fig. 356. Fig. 357. 

its velocity is in a direction making, let us say, an angle S9 with 
its old direction. In Fig. 357 let o b repres^ent to scale the velocity 
at p ; and let o c, equal in length, but making an angle co b =z d9, 
with o B, represent the velocity at q. Now, in vector addition, 
OB + BC = OC; so that if Si is the time taken by the body to 
move from p to q, in that time there has been the lateral change of 
velocity BC. bc=:ob.^0 more and more nearly as 89 is made 
smaller and smaller, or b c = v . £9. Hence the lateral accelera- 
tion, which is lateral change of velocity divided by the time of the 

The academic person may be quite right to stick to his conventions as to 
force, because he never has to think of the medium (string or other) through 
which forces are exerted when he is working his problems. The engineer is 
compelled to deal with the larger question ; he very wisely converts all his 
dynamical problems into statical problems, and all his forces are always 
balanced. 



APPLIED MECHANICS. 605 

change, is v . - or w — . - = v- - z= y2 ^ curvature, because 

00 OS oC OS 

d9 
our definition of curvature is -^. In a circle the curvature dO/ds 

happens to he the reciprocal of the radius. Hence we speak 
of this acceleration towards the concave side of a body's path as if 
it were moving in a circle (curiously enough, the fact seems always 
to he forgotten that the path is usually not a circle, nor even 
the very smallest arc of a circle) of radius r with, a centripetal 
acceleration v-/r.* The centripetal force causing the change of 
motion is the mass multiplied by this. The engineer is usually 
concerned with the equal and opposite force which the body exerts 
upon the constraints, and calls it centrifugal force. We see that 
it is mv-lr ovvia^r or wrn-l29S1, where vi is the mass of the body 
or w is its weight in pounds at London, r the radius of curvature 
of the path in feet, v the velocity in feet per second, a the angular 
velocity in radians per second, n the number of revolutions per 
minute. Observe that if masses vi-^ and m^ are attached to the 
same shaft at distances r-y and o'^ from the axis, their centrifugal 
forces are in the proportion of m-^^ r^ to vi^ r^ ; if we have half the 
mass at twice the distance, we have the same centrifugal force. 

If a body changes in its speed r, and so has an acceleration in 
the direction of its path, its total acceleration is the vector sum of 
the two accelerations, and the resultant force acting on the body at 

any instant is the resiiltant of m. — along the path and m — 

at right angles to and in the plane of the path. 

Example. — A body of w lbs. is moving along a curve with a 
velocity v and an acceleration a in the direction of motion ; the 
radius of curvature of the curve is r. What is the total accelera- 

tion and the force causing it ? Here we have — and a two accelera- 

tions at right angles to one another. The answer is, an acceleration 

V -75 + a^ in the direction makino- an angle 9 with the direction 

,,.1 o o 

of motion, where tan. = — • Multiply by the mass w -^ 32-2, and 

we have the total force. 

501^ Centrifugal Force in Belts or Ropes. —In Fig. 356 let p a be 
a small portion of a flexible body oiw .5s lbs. ; its centrifugal force 

* To keep in our minds the fact that all motion is relative, we ought to 
remember that, relatively to the body, other bodies have a centrifugal 
acceleration. The words " centrii)etal " and "centrifugal" are technical 
terms now; their origins seem to have had something to do with the notion 
that a curve has millions of centres. It is all very well for the mathe- 
maticians to speak of a small i^art of a curve as being an arc of a circle, but the 
engineer knows that it is only in this matter of curvature or rate of change 
of d with s that it is like an arc of a circle, for he knows that complete 
information about tiie very, very smallest portion of any curve implies a 
complete knowledge of the Avhole curve. 



606 



APPLIED MECHANICS. 

is - Ss . 1, or - 5s . ?,-2 . ^, or ^ v^ . dO. If the tension is t lbs., 

we know from Fig. 357 that t . d9 = - t"- . SO, so that t = ^ ^•2. 

ff g 

If a is the section in square feet, and / is the tensile stress per 
square foot, and w^ is the weight of 1 cubic foot of the material, 

w = aw^, T = <?/; so that/= -^ f^ being independent of the radius 
of the path. 

Example. — What velocity will produce a tensile stress of 
3,000 lbs, per square inch in the thin rim of a cast-iron pulley? 
Here/ = 3,000 x 144, and w^ = -26 x 1,728. Hence 

26 X 12 „ /32-2 X 3,000 



^^^ -26 X 12 „ / 

,000 = — ^ f2 or y = v 



32-2 "^ -26 X 12 ' 

t? = 175 feet per second ; 
or, in engineers' language, a Telocity of 10,500 feet per minute 
will produce what is usually taken as the working tensile stress in 
cast iron. 

502. If the plane of the path alters, if .the plane rotates about the 
tangent to the path through the angle S0 in the distance 55, then 
d(pjds is caUed the tortuosity of the path. When a body moves 
in a tortuous cui-ve it has acceleration clvlclt along the path and 
v^jr, a centripetal acceleration in the plane of the path, or the plane 
of cmwature, as it is called. And if a model made of three very 
short pieces of wire, o s, o r, and o n, be made, the angles between 
OS, OR, and o n being right angles, and if we conceive o s to keep 
parallel to the path 55, if o r keeps pointing to the centre of curva- 
ture, then the angles turned through about the axes o n and o r per 
unit length of the path represent the curvatiu'e and the tortuosity. 
A student ought to make a model of a curve with wire and let a 
little frame like this slide along it, and study the matter for 
himself. A spiral path in which the curvatm-e and tortuosity are 
constant is particularly interesting. If we refer the position of a 
particle to thi^ee axes of reference, its total acceleration at any 
instant is compounded of the three d^xjdfi, d'^yjdt'^, drzjdt'^. The 
three comjDonents of the resultant of all the forces which are acting 
are in times these, if m is the mass or inertia of the particle. 



THE BALANCING OF MACHINES. 

503. If a wheel is fixed eccentrically on its shaft, or if to a 
shaft there is attached any object whose centre of gravity is 
not exactly in the axis when the shaft rotates, centrifug'al 
force causes pressures on the bearings of the shaft wdiich 
are always in the direction of the centre of gravity of the 
rotating mass. In this case there is said to be a want of 
balance. If yon wish to observe the effect produced by such 
want of balance, mount an axle to which a wheel is keved on 



APPLIED MECHANICS. 607 

any support which is not very lirm ; fix a small weight 
on one of the arms of the wheel, and rotate it rapidly. 
You will find that, even if the weight is small, surprising 
effects are produced, and show themselves in a shaking of 
the supports ; and the evil effects are four times as great 
at 200 revolutions per rainute as at 100 revolutions per 
minute. Centrifugal force is proportional to the mass of 
a rotating part multiplied by the distance of its centre of 
gravity from the axis of rotation, multiplied by the square of 
the number of revolutions per minute. 

504. If a number of bodies are attached to a shaft and are 
whirling round with it, each of them at any instant exerts a 
force on the shaft which can be calculated, and the resultant 
effect on the two bearings may easily be determined, just as 
easily as in the static problem of Art. 99. If the axis of 
rotation passes through the centre of gravity of all the 
rotating parts, the pressure on one bearing is equal and 
opposite to the pressure on the other; and by properly 
placing the masses, the pressure on either bearing may be 
reduced to nothing. Thus it is evident that when two masses 
are directly opposite to one another on a shaft, their cen- 
trifugal forces may be made to balance one another. When 
not opposite they cannot be made to balance, but two masses 
may balance one which is directly opposed to the resultar.t 
force of the two. When there is no pressure on either of 
the bearings, so that there is no tendency to change the direc- 
tion of the axis, it is said to be the loermanent axis of the 
rotating masses. All axes of rotation in machines ought 
to be permanent axes. When this is the case in a rotating 
machine, and it is suspended by ropes and made to work, there 
are no visible oscillations. 

505. The balancing of a machine consists in adding masses 
in such positions, or re-arranging the positions of the 
existing masses so that the centrifugal forces due to their 
rotation are just able to balance the otherwise unbalanced 
forces which act on the various shafts. The student will find 
that the study of one problem in balancing will make him 
familiar enough with the method of calculation for its applica- 
tion to almost any other case which is likely to occur in 
practice. The most usual case for the student to take up is 
that of the locomotive engine, because want of balance in the 
locomotive is capable of producing very serious effects indeed. 



608 



APPLIED MECHANICS. 



Fig. 35 S shows an electromotor driving a shaft on which a 
number of discs are keyed. Weights may be fastened on 
these discs : the want of balance is evident when the shaft 




Rg.358. 

rotates, and students will find it easy to illustrate how the 
shaft may be balanced by other weights properly placed. 
Thev will see that when motions are merely rotatory we can 
always have a perfect balancing of machinery. 



APPLIED MECHANICS. 609 

506. Sxamplc. — It has beer., shown by experhiient that the appli- 
cation of suitable balance weights is attended by a sensible reduction 
of resistance on railways at high siDeeds. Locomotive engines 
unbalanced cannot attain as high speeds as when balanced, with 
the same consumption of fuel. There are two separate sets of 
unbalanced forces acting on the crank shaft of a locomotive. (1) 
The centrifugal force of the crank, crank-pin, and as much of the 
connecting-rod as may be supposed, roughly, to follow the path of 
the crank-pin (say one-half of it). The mass or weight of each of 
these multiplied by the distance of its centre of gTavity from the 
axis, divided by the length of the crank, gives the mass which, on 
the crank-j)in, would produce the same centrifugal force. Let this 
weight be called tv lbs. In designing engines, we consider half the 
connecting-rod to act as if collected at the crank-pin , the other half 
to be moving with the piston. At the end of the stroke, when the 
horizontal component of the centrifugal force is greatest and the 
vertical component vanishes, the horizontal pressure on the axle 
caused by the centrifugal force is 

10 v- 'W /27rK^i\'2 1 

32^ • R °^' 32^ i -60- ) K °'" ''^■'"'■" -^ -"'"7' 
K T)euig the length of the crank in feet, and n the number of 
revolutions per minute. (2) We have the force due to the mo- 
mentum of the reciprocating mass, including piston, j)iston-rod, 
slide, and the second haH of the connecting-rod. The loss of 
momentum is most rapid just at the end of the stroke ; and as loss 
of momentum per second is what we call force, the force acting 
on the axle at the end of the stroke due to this cause is easily 
found, and proves to be 

w R «2 ^ 2,937 

where w is the weight of the total reciprocating mass. 

Now a weight ^v\, or weights whose sum is Wi, may be placed 
on the driving-wheel or wheels at a distance r from the axis, such 
that the centrifugal force of Wi may be equal to the sum of the 
above forces. This leads to 

ivir = w R + WB.; 

and if we assume any distance, ;•, we can calculate the balance 
weight or weights, wi. 

607. Now, for the axis to be permanent in inside -cylinder engines, 
Wi must be di-sided into two parts, one for each wheel, inversely 
proportional to the distances of the wheels from the cranh. For 
outside- cylinder engines we get balance weights for the two wheels 
whose difference is Wx, and they are, as before, inversely proportional 
to the distances from the wheels to the crank in question. Hence, 
a consideration of each cylinder gives two balance weights, one 
usually much smaller than the other. As the cranks are at right 
angles, the balance weights ought to be 90° apart on each wheel. 
Instead of using these two, we can use one weight placed between 
their positions, so that its centrifugal force is the resultant of 
theirs. Thus, if we found 20 lbs. and 6 lbs. for the two placed at 
the same distance from the axis but 90* apart, make o a equal 20, 



610 APPLIED MECHANICS. 

and o B, at righ.t angles to o a, equal to 6 according to any scale ; 
complete the parallelogTam, and o c rejiresents on the same scale 
the weight which will replace them. It ought to he placed at just 
the same distance from the axis as they were supposed to be 
placed; and in position it makes the angle ago with the larger 
weight. In this case it wiU he found that 20-88 Ihs. placed 18-3° 
from the position which the weight of 20 lbs. might have occuijied 
will be required to replace the two. 

508. It often happens in outside- cylinder engines that the 
distance from one wheel, or rather from the centre of gTavity of a 
balance weight, to the crank, is so little that the corresponding- 
weight for the other wheel is very small, and may oven be neglected. 
In inside-cjdinder engines it will be fomid that, whereas the cranks 
are at right angles to one another, the balance weights on the two 
wheels on the oi^posite side of the axis to the cranks are often only 
50° apart. In inside-cylinder engines with coupled wheels the 
outside coupling rods and cranks ore usualhj made to balance the inside 
moving iJarts. These engines work very smoothly indeed. Outside- 
cylinder engines with coupled wheels are "\^ery unstable, from the 
use of small wheels requii-ing very rapid revolution of the crank 
axle ; from the cylinders being farther apart than usual, so that 
the coupling-rods may have room, and from the number of 
reciprocating parts being increased. The conditions seem to admit 
of no remedy for these defects. The balance weight ought to be 
distribiited over two or three of the spaces of the wheel,, that the 
tire may not be unduly strained. 

509. We have, then, the following easy, approximately correct 
rules for locomotives : — If r is length of crank, r the distance of 
centres of gravity of every balance weight from centres of wheels, 
e the distance apart of the centre lines of cylinders, d the distance 
apart of the wheels or centres of gravity of the balance weights, 
w the total weight of crank (referred to the pin), pin, connecting- 
rod, piston, sHde, and piston-rod, a the angle which the position of 
centre of gTa-\'ity of balance weight makes with near crank : 

(1) Inside-cylinder engines -with unco-upled -wheels. Each 



balance weight = -^r-r-, .s^/ 2 (^ -f 2 e^, tan. a = -r-. — ; so that a is 
z dr ct -T 6 

easily obtained from a book of tables. 

(2) Outside -cylinder single engines with uncoupled wheels. 

Each balance weight = — ;-, a = 180° ; so that in this case the 

balance weight is placed exactly opposite to the crank. 

(3) Inside -cylinder engines -with wheels coupled. Find by 
rule (1) if the weight of the coupling-rods, etc., is too great. If so, 
let counter weights equal to the difference be placed opposite the 
outside cranks. If too small, the difference must be made u]3 with 
balance weights, as in rule (1). The positions of the outside 
cranks are fotmd by rule (1). 

(4) Outside-cylinder coupled engines. Find revolving weight 
of coupling-rods, etc., for each wheel. Also find sum of the weight 



APPLIED MECHANICS. 611 

of the piston, rod, slide, and half connecting-rod. Divide this 
latter among the wheels, adding the given revolving weight 
already on them. Let this he used on each wheel according to 
rule (2). 

510. We have dwelt upon these practical riiles for balancing in 
locomotives, hecause they give good illustrations of centrifugal 
force. But the student ought clearly to see that it is only when, a 
hody rotates about an axis that we can exactly balance the forces. 
A body with reciprocating motion can only be balanced by 
another body with a reciprocating motion ; and hence it is that, 
after much expense and quarrelling with persons complaining about 
the vibration of their houses, many electric lighting companies have 
discarded reciprocating steam-engines, rex)lacing them with steam 
turbines. In Arts. 427 and 429 we give the principles on which the 
subject may be studied. I give practical exam23les of their use in 
my book on steam, gas, and oil engines. In that book I give the 
more exact constructions, which are so commonly taught now to 
advanced students, to find the forces at the ends of a connecting- 
rod for any position. It seems never to strike a student or his 
teacher that such elaborate calculations may possibly not give a 
very different result from what they may obtain by the simj^le 
assumption like what I bave stated above. One of mj students 
has made the comparison. His three weeks' constructions to find 
the forces acting on the frame of a Willans engine, the turning- 
moment on the crank shaft, etc., nowhere differ appreciably in 
theii' results from those obtained by assuming that half the mass 
of the connecting-rod has the motion of the cross-head, and the 
other half that of the crank pin. The very much more important 
matter, the effect of the motion of the rod upon its strength as a 
laterally loaded strut, seems never to engage much attention. 

EXERCISES. 

L A crank pin 4 inches in diameter and 6 inches long has to be 
balanced. If the length of crank be 9 inches, and the balance weights 
are placed directly opposite each crank arm, find the weights, the centre 
of gravity of each being 6 inches from the centre of the crank shalfc. 

Ans., 16-24 lbs. 

2. A shaft is in balance under the action of three weights, one of 300 
lbs., at a distance of 12 inches from the axis; another of 100 lbs., at a 
distance of 20 inches from the axis, on the opposite side, and 30 inches to 
the right of the first. How much must the third weight be, and where must 
be its position along the shaft, if its distance from the axis is 28 inches ? 

Ans., 57 '1 lbs.; 37.5 inches to left of first weight. 

3. In a locomotive the distance between the centre lines of the 
cylinders is 27J inches. Balance weights are fixed at a horizontal 
distance apart of 59 inches, the centre of gravity of each describing a 
circle of 55 inches diameter. If the weight of the reciprocating masses 
for each cylinder be 400 lbs., and the stroke be 25 inches, find the 
position and magnitude of balance weights to counteract the horizontal 
and alternating force and couple. 

Ans., At 160° with near crank; 142 lbs. each. 



612 APPLIED MECHANICS. 

4. In a steam-engine the piston at tlie beginning of its stroke is 
exposed to a total effective steam pressure of 2,000 lbs., but the inertia of 
the piston is such that the thrust of the piston-rod is only 1,600 lbs. 
The speed of the engine is now raised until it becomes half as great again 
as before, while the steam-pressure is unchanged. What is the thrust of 
the piston-rod ? Ans., 1,100 Ihs. 

5. An engine is making 150 revolutions per minute. What is the 
acceleration of the piston at the commencement of each stroke, the 
connecting-rod being 4 feet long and the crank 9 inches ? 

Ans., 223 ; 153. 

6. The weight of the reciprocating parts of a steam-engine is 
equivalent to 3 lbs. per square inch of the area of the piston. If the 
length of crank be 9 inches, find how much the initial effective pressure 
is reduced by the inertia of the reciprocating parts when the crank makes 
70 revolutions per minute, the obliquity of the connecting-rod being 
neglected. Ans. 3*8 lbs. per square inch. 



613 



CHAPTER XXYIII. 

SPRINGS. 

511. Any contrivance which can store energy as strain 
energy, and give it out again readily, is a spring. Hence, any 
tie-bar or strut, any beam — in fact, any object whatsoever is a 
spring. The term is, however, generally applied only to such 
objects as can be changed in shape very much without fracture. 

A tie-rod of indiarubber can be stretched to eight times its 
old length, again and again, without hurt ; whereas a tie-rod 
of the best steel can only be stretched to -sV^h of its old 
length, again and again, without hurt. 

Hence the indiarubber tie-rod or a strut may be called 
a spring, just like the spiral spring ; but it would not be right 
to speak of a tie-rod or strut of steel as a spring. The differ- 
ence is, however, only one of degree, and indeed, a mine cage 
suspended by a steel rope half a mile long vibrates up and 
down just as if hung from any ordinary spring. 

512. Springs are almost always used as reservoirs of energy 
— that is, as hydraulic accumulators are used, or fly-wheels of 
steam-engines, or cisterns of water, or electric accumulators. 

The mainspring of a clock or watch takes a store of energy 
in winding-up, and gives it opt gradually for about twenty- 
four hours in unwinding. A how gets gradually a store of 
energy, which it gives out rapidly when the arrow is set free. 
A buffer-stop spring stores up all the kinetic energy of a train, 
and the stiffer it is the more quickly will it store the energy, 
and therefore the more suddenly will the train be brought to 
rest. Ordinary buffer springs are continually storing up and 
giving out energy, equalising only gradually the velocities of 
the two railway carriages, so that if one gets a sudden change 
of velocity, the other shall only be affected gradually. In the 
same way, when any two objects have a springy connection, if 
one of the objects alters its velocity suddenly, the other alters 
its velocity in consequence only gradually. These are cases in 
which a spring is used to prevent shocks, or, as we may put 
it, a spring is used to lengthen the time of a blow, and there- 
fore to diminish the average force of a blow. 

513. Now there are several distinct cases here to consider. 



614 APPLIED MECHANICS. 

I. A body of mass b, moving with velocity v , overtakes a 

b<xly A, moving with velocity v . The buffer between them is 

strained until they move with the same veLDcity. If the 
common velocity then is v, 

« h - '■) = ^ (^' - \) 

B I" + A r 



A -f B 

Tl:e energy now stored up in the spiing is — 

iBcl + ^ Av^ - 1 (A -f b)i;2. 

^ b - a - ' 

In case the bodies b and a are themselves elastic, they them- 
selves,, to a greater or less extent, act as buffers, and less 
energy is stored in the buffer itself. Also as changes of shape 
are usually accompanied by friction, some of the energy is 
wasted. If no energy were wasted, the two objects would 
keep going faster and slower relatively to one another, as the 
spring was compressed and extended, and to some extent this 
does go on after the blow : but these vibrations are usually 
rapidly stilleii as the energy is wasted in friction, partly, as 
we have already said, in the buffer spring itself, and partly by 
friction opposing generally the motion of the bodies. 

II. If one of the b<xlies, a, is fixed to the earth, then the 
mass of A mav be resarded as infinite : v and v in the above 

calctdation become 0, but the same reasoning applies as before. 

III. Two carriages, a and b, are at rest connected by a 
spring. A is made suddenly to move through a distance a. 
Xow if the spiing were infinitely stiff, b would just as suddenly 
move through the same distance a. Xs, the spring is less and 
less stiff, B moves over the distance a with less and less 
suddenness, because the kinetic energy which must eventually 
be given to b is suddenly stored in the spring, but is only 
gradually given to b by the spring. If there is no friction, b 
wdl be left vibrating. If there is no friction opposing b"s 
motion, but there is friction due to change of shape of the 
sp)ring, B will vibrate and gradually come to rest. 

If there is friction opposing Bs motion, but there is no 
fiiction in the spring, the store of energy in the spring is 
cn-eater than before, b must, of course, come to rest. With 
friction in the spring, b will more rapidly come to rest. 



APPLIED MECHANICS. 615 

In all these cases, b comes to rest at the distance a from 
its old position. 

lY. As in the last case — but A moves suddenly through a 
distance a, and after a short time T is moved suddenly back 
to its old position. 

(1) If T is exceedingly small, b does not move. 

(2) If T is great, b moves as described in Til., and repeats 
its motion in the reversed direction. 

(3) If T is neither too great nor too small, b has a motion 
intermediate between the (1) and (2) described motions. 

(4) If a's motion is quickly vibrating through the ampli- 
tude a, B gets a vibratory motion of the same period ; but 
superimposed on this is the natural vibratory motion of a 
period which depends on the stiffness of the spring and the 
mass of b: the natural vibrations will die away if there is 
friction. 

514. It is well worth while for a student to illustrate this last 
case by means of a model. The crank q. Fig. 359, is turned 
round regularly ; if this is done ^ 

by hand, a fly-wheel ought to be 
used to give steadiness of motion. 
By means of a connecting-rod, p 
gets an up and down motion 
which, diminished in the ratio 
o a/o p, is given to a ; this creates 
a forced vibration in b. The 
natural vibration of b, when _____ 
A is not moving, ought previ- Fig. 359. 

ously to have been studied — (1), 

when there is very little friction, and b goes on vibrating with 
nearly the same amplitude for a long time 3 (2) when there is 
fluid friction, b vibrating in a vessel of water. In this case 
there is reaJly a change of inertia difficult to calculate; the 
water being set in motion. The auiplitude of b's motion gra- 
dually diminishes because of friction. When a suddenly begins 
to vibrate, b may have a large natural vibration, but this 
gradually gets destroyed by friction, just as if there were no 
forced vibration. We shall now speak of the forced vibration 
only, and assume no friction. Let A vibrate with / times the 
frequency of the natural vibrations, and let a's amplitude be 1. 
Then the amplitude of b's motion will be as follows : when 
there is no friction b moves synchronously with A, so that B 




616 



APPLIED MECHANICS. 





.— 1 




O 










o 


■ — ■ 




' 




~! 




IJ. 














' 


1 




-- 














■r> 








-r- 














" 


1 












o 


1—1 


t- 


^ 


1 




_^ 






9 


i 


'"' 


7 


^ 


^ 


o 


o 


'^ 


1 


^ 


8 





-"7^— 




















O 




O 


ac 


c< 


Ci 


vb 




oi 




2? 


t- 




■Cl 


cb 




' ! 




•O 




















■—1 




1 ^ 




O 


C5 


ca 


' 


' kb 




1 ^0 




1 t^ 


GO 


! *^ 




oi 




CO 




CO 


O 






' -^ 




^ 


— 


■z. 




^ 




ji 








-■ 1 




c 




a 




'O 




















"5- 




-< 



is at the top when a is at the top, or else 
(and this is indicated wdth a — sign) b is at 
the bottom when a is at the top. 

To prove tliis, let w be the weight in 
lbs. of B, so that its mass is w/y. As 
before, a better approximation to accuracy 
is obtained by letting w iaclude one- 
third* of the weight of the spring. Let 
the spring be sncb that a force of 1 lb. 

w 
elongates it A feet. Then - x the accelera- 
tion is the force in the spring; and if ar is 
the distance in feet of the body below its 

position of equilibrium, - A — . -tj^ is the 

extra elongation of the spiing when there is 
this acceleration. But if b is x feet below, 
and if A is 2/ feet below their positions of 
equilibrium, then x — y \a this extra elonga- 
tion, and henc :■ 



, w drX 

■ h — ^ 

9 dt- 

cPx ^ a 



y 



.(1), 



Lettino- 



J_ 
wA' 



(2). 



we see that n divided by 



2 T is the natural frequency of vibration. 
As in Art. 19, if we introduce our force of 
fiiction equal to h times the velocity, using 
2 F for hgl^i, we have 



(Px ^ dx 

^ 2 F — 

df- ■ dt 



n^ = n^ 



(3). 



By putting y = a sin. qt, it is easy to find 
the forced vibration. For simpHeity, if we 
let F = o, then 

a 

X = -z ^— "i vSin. ot. 

I - ^- n- ^ 

Table XTT. is obtained by letting q/n be called/. 

!N"ote in the model and from the tabled 
numbers that when the forced frequency is 
a small fraction of the natural, the forced 
vibration of b is a faithful copy of the 
motion of the point of support a ; the spring 

* Let the student prove that j. of the weight of 

the spring is to be taken and not | of it. 



APPLIED MECHANICS. 617 

and B move like a rigid body. When the forced frequency 
is increased, the motion of b is a faithful magnification of 
a's motion. As the forced gets nearly equal to the natural, 
the motion of b is an enormous magnification of a's motion. 
There is always some friction, and hence the vibration cannot 
become infinite. When the forced frequency is greater than 
the natural, B is always a half-period behind A, being at the 
top of its path when A is at the bottom. When the forced is 
QQany times the natural, the motion of B gets to be very small ; 
it is nearly at rest. 

515. Men who design Earthquake recorders try to find a 
steady point which does not move when everything else is 
moving. For up and down motion, observe that in the last 
case just mentioned b is like a steady point. 

When the forced and natural frequencies are nearly equal 
we have the state of things which gives rise to resonance in 
acoustic instruments ; which causes us to fear for suspension 
bridges or rolling ships. 

It is obvious, then, that the simple statement of the 
problem, " How do springs prevent shocks ? " presents, when 
we consider it very carefully, many quite different problems. 
It is worth while observing carefully the up and down motion 
of the body b of a waggon on the street when the wheels A 
go up and down over the stones. But we can study the sub- 
ject better perhaps in the model. We see that although a 
spring connection between A and b does prevent shocks, the 
motion of b may be dangerously great. Thus, for example, 
when an earthquake occurs it does not always do to merely 
have an elastic connection between the ground and the house, 
as the earthquake leaves a Japanese house vibrating sometimes 
60 much as to give quite a sea-sick feeling to the inhabitants. 

We see that in all cases, unless there is friction opposing 
vibration of the body b — whether this friction exists in the 
parts of the spring itself, or more directly opposes the motion 
of B — there will be vibrations, sometimes dangerously large in 
amplitude. When by means of the spring connection we seek 
to diminish shocks, friction may be introduced by some dash- 
pot arrangement, which may consist of a porous piston moving 
in a cylinder filled with water or oil ; or the piston may be 
solid, and there may be a pipe and cock connection between 
the opposite sides of it. Here the friction is mostly fluid 
friction. If it were all fluid friction, there would be no 



618 APPLIED MECHANICS. 

opposition to iiiotion — there would, in fact, be no friction — if 
the motion were slow ; the friction is greater and greater as the 
motion is quicker and quicker. The friction between the plates 
of a carriage spring, which rub together every time the spring 
is changed in shape, is solid friction, which, if anything, is 
probably greater for slow motions than for quick motions. 
In all cases when vibrations are to be stilled it is better that 
the friction should be of the nature of fluid friction, but this 
is not always convenient. Its effect in stilling vibrations may 
very readily be studied in the laboratory. It will be found 
that by adjusting the cock of this dash-pot we can vary at 
will the rate of stilling of the vibrations of even very large 
masses, so that after a shock through the spring the body B 
may vibrate for a long time before it comes to rest, or it may 
come to rest after one slow lurching motion only. Now, it is 
necessary to understand that a dash-pot arrangement, or any 
other arrangement for introducing fluid friction, will not 
affect the static law of the spring in any way. It may be 
introduced on spring balances which are required to measure 
forces accurately, for example. But the sort of friction we 
find in carriage springs is very different. Here the plates of 
which the spring is made rub on one another, and there is 
solid friction, and such a spring as this cannot be used for 
spring balances. The law of the spring is altered ; we cannot 
depend upon the spring for measuring forces if there is any 
place where rubbing of solids takes place. Such springs are 
never used for purposes of measurement ; they are only used 
for preventing shocks. 

516. Springs are of many diff'erent forms ; . they are used as 
small as balance springs in watches, and they are sometimes 
so large as we see them in some buffer and locomotive springs. 
The following hst is not given as by any means an exliaustive 
one : — 

I. Cylindric spiral springs, subjected to axial loading. — In 
these it will be found that the wire, whether round or nearly 
square in section, is twisted Hke an ordinary revolving shaft 
transmitting power, and the strain is one of torsion in the 
material. 

Examples. — Most forms of spring balance and dynamo- 
meters ; the springs of indicators ; many railway carriage and 
tram-car springs ; safety-valve springs for marine and loco- 
motive boilers. 



APPLIRD MECHANICS. 619 

II. Cylindric spiral springs, subjected to a torque about 
the axis : — In these it will be found that the material is sub- 
jected everywhere to bending. 

Examples. — The balance springs of the best chronometers ; 
the springs used as elastic joi'^ts between two lengths of 
small shafting. 

III. Flat spiral springs, subjected to a torque about the 
axis. — In these it will be found that the material is subjected 
to bending. 

Examples. — The main and balance springs of watches and 
many clocks ; the springs used in nearly all contrivances 
which require to be wound up, or wherever a large reservoir 
of energy is required in a small space. 

IV. Nearly straight strips of material subjected to bend- 
ing. — These are used in a great variety of cases, sometimes in 
one piece, as in the limb of a tuning-fork. 

V. More or less flat, or corrugated, circular, or of other 
outline, fixed or only supported at three or more points at the 
edges. 

YI. Indiarubber springs, usually subjected to merely ten- 
sion or compression, now being used largely for tram-cars. 
In all the above cases there is an absence of solid friction. 

VII. The Ayrton-Perry spring, used in indicating the 
amount of a force by a large relative rotation of a pointer. — 
It will be found that the material in these springs is subjected 
to a combination of bending and twisting strains. The Ayrton- 
Perry twisted strip, in which a very smaU elongation is accom- 
panied by great relative rotation. 

VIII. More or less straight strips of material subjected to 
bending (like IV.), a number of pieces being used in one 
spring, these pieces rubbing on one another. When bending 
occurs, this introduces solid friction. 

Exatnples. — Locomotive, waggon, and carriage springs, 
and nearly all the large springs used for minimising shocks, as 
buffer-stop springs. 

IX. Gases in closed vessels, the volume of which may be 
altered, as in the air-chamber of some force-pumps ; the 
cylinder and piston air-spring. 

It is, however, obvious that the functions of springs 
cannot be specified in a fe\f words. Springiness comes in 



620 APPLIED MECHANICS. 

1 

usefully in the packing-rings of pistons, and in packing 
generally, to produce a good fit between pieces which rub 
on one another ; in spring split rings, as washers, which 
prevent a nut from becoming loose ; when a springy piece of 
material is used between two more rigid pieces which are 
bolted together, to give a uniform bearing without undue 
strains in the nuts. When, in fact, we discuss the elasticity 
of material generally, we see that everything in nature is a 
spring, and performs most of its functions in nature by means 
of this elastic property. 

517. The main uses to which springs are put are these : — 

1. Lengthening impacts, so as to diminish the forces of 
blows, and therefore absorbing, and in these cases usually also 
dissipating energy. 

2. Regulating motion — that is, preventing large fluctua- 
tions in speed in driven pieces of machinery. This is not a 
common use of springs, because of the waste of energy. 

3. As reservoirs of energy. 

4. Regulatmg, as in watches and clocks. 

5. As measurers of force. 

6. As measurers of distance. 

7. As measurers of angles. 

518. The Best Materials to Use in Springs. — A spring's 
usefulness depends primarily on its being a reservoir of energy. 
In the first two cases of the preceding table this capacity for 
storage of energy, and of course cheapness and ease of manu- 
facture, ought to settle for us the material of which a spring 
should be made. In the other cases we must also consider 
the question of perfect or imperfect elasticity and viscosity of 
the material. First, then, as to the energy which may be 
stored. 

The energy which must be given to distort a spring before 
it takes a permanent set is called its resilience. 

Now, it will be shown in Art. 535 that in all springs sub- 
jected to bending, as springs of classes 2, 3, 4, 5, 7, and 8, the 
resilience per unit volume of the material depends upon the 
resilience per cubic inch of the material when subjected to 
compressive or tensile force, and this is f ^ -^ 2 E in inch- 
pounds, where / is the greatest tensile or compressive stress 
which the material will .stand without taking a set, and e is 



APPLIED MECHANICS. 



621 



Young's modulus of elasticity. The following table shows 
the value of this constant for various materials : — 



TABLE XIII.— Spring Materials 


Subjected to BE^ 


DING. 




/ 


E 

lu Millions of Pounds 
per Square Inch. 


/2^2e 


Wrouglit Iron 


24,000 


29 


10 


MHd Steel 


35,000 


30 


20 


JVIild Steel, Hardened 


70,500 


30 


83 


Cast Steel, Unliardened 


80,000 


30 


107 


Cast Steel, Hardened 


190,000 


36 


50] 


Copper 


4,300 


15 


0-62 


Brass 


6,950 


9-2 


2-62 


Gnn Metal 


6,200 


9-9 


, 2-00 


Phosphor Bronze 


19,700 


14 


13-85 


Glass 


4,500 


8 


1-26 



The numbers in Tables XIII. and XIY. are subjected to very 
considerable variations, especially in the cases of copper, 
brass, gun metal, phosphor bronze, and glass. Indeed, in our 
opinion, definite statements as to the values of /^ -^ 2 e, or 
fy -r- 2 N, ought not to be made, until careful experiments 
have been made on such varieties of these materials as are 
actually used in spring-making, and this has not yet, we be- 
lieve, been done. We have taken the values of y, e, and n from 
Table XXII. In the best spiral steel springs, for example, 
the value of f^ (the proof shear stress) has been found to be 
rather 60,000 or 70,000 lbs. per square inch than the 145,000 
given in the table. 

It will be shown in Art. 535 that in all springs where the 
material is subjected to twisting merely, as springs of Class 1 
— the most important class, probably, for the use of mechanical 
engineers and instrument makers — the resilience per unit 
volume of the material depends upon the value of 

f{ 

2.n' 
where /j is the greatest shear stress which the material will 
stand without taking a set, and N is the modulus of rigidity of 
the material. The following table shows the value of this 
constant for various materials : — 



622 APPLIED MECHANICS. 

TABLE XIV. — Matekl^ls for Cylindhic Spiral Springs. 





/i 


N 

In Millions of Pounds 

per Square Inch. 


/i2-5-2n 


Wrou2:lit Iron 


20,000 


10-5 


19 


]\lild Steel 


26,500 


11 


32 


Mild Steel, Hardened 


53.000 


11 


128 


Cast Steel, Unhardened ... 


64.000 


11 


186 


Cast Steel, Hardened 


145.000 


13 


809 


Copper 


2.900 


5-6 


0-75 


Brass 


5.200 


3-4 


4-00 


Gnn Metal 


4,150 


3-7 


2-33 


Pliosplior Bronze 


14,500 


5-25 


20 



The numbers given in Tables XIII. and XIV. are supposed to 
express, then, the actual relative values of the various 
materials for spring-making. 

It will be observed that hardened cast steel is very much 
better than any other material for spring-making ; hardening 
makes it five times more valuable. It is about 35 to 40 
times more valuable than phosphor bronze ; more than 40 
times more valuable than wrought iron (which is not so good 
as phosphor bronze). The fact that phosphor bronze makes 
probably the best non-magnetic material for siDrings has 
been known to me for fifteen years. I tested tliis result 
by a great deal of experimenting with various materials. 
But this is not the only virtue of phosphor bronze. In quite 
a remarkable degree it is free from many of the vices of other 
metals — sub-permanent set after small loads, efiects due to 
fatigue, etc. It is worth while to mention that by the nature 
of the process of manufacture, the material may have initial 
strains in it ; before applying it in an instrument it ought to 
receive a considerable set in the direction in which it will 
most usually be strained. Phosphor bronze springs in my 
electrical instruments receive a set from a load which is six 
times as great as the greatest load ever applied to the spring 
when it is in use. 

The numbers in the tables give us guidance, but we must 
also consider special conditions. The hardening and temper- 
ing of steel require great care ; so great is this that we may 
almost say that there is only one steel spring maker m the 



APPLIED MECHANICS. 



623 



whole of England. Now, phosphor bronze and brass and 
copper receive their greatest hardness by drawing through 
dies or rolling. They can, in fact, be hardened very uni 
formly in the cold state quite readily, and springs of them are 
easy to make. Again, although tempered steel has usually an 
oxide of iron to protect it, and a soft iron spring of any kind 
can also be given such an oxide by Barff's process as a cover- 
ing, yet, on the whole, steel and more especially iron springs 
are much more subject to rust when exposed to a damp atmo- 
sphere than copper, brass, or phosphor bronze. Again, in 
certain electrical instruments where springs are used, steel 
and iron must not be used because of their magnetic pro- 
perties, and in other measuring instruments the properties 
catalogued in the tables may not always be all-important. 

519. The following property is not of importance in springs 
used to prevent shocks, as in buffer and carriage springs : — 

When a spring is loaded with even a small load it may 
continue to lengthen axially slowly if the load is kept on ; and 
afterwards, when the load is taken off, it may not immediately 
shorten to its original length, but needs time. This is usually 
called sub-permanent set, and is greater with greater loads. 
When such a spring is unloaded after it has experienced loads 
of various amounts and various periods of rest, it will not 
usually go back to its old length, but will slowly undergo 
slight shortenings and lengthenings of various amounts de- 
pending on its previous experiences. I sometimes call this 
property the " creeping " property of the 
material. Professor Ayrton and I have 
written a paper concerning this property. 
A material possessing much of it is quite 
unsuitable for the springs of measuring 
instruments. 

520. Spiral Springs. — A spiral sprnig is 
a wire, or rod, or strip of any constant or 
varying section (we shall always speak of 
it as a wire, of whatever size or shape its 
section may be), coiled so that the centre 
line of the wire lies everyv/here on some 
surface of revolution. In most cases the 
wire is wound on a cylindric surface, the 
winding being perfectly regular — that is, the 
angle made by the centre line of the wire, 




Fig. 360. 



624 APPLIED MECHANICS. 

with a plane at right angles to the axis of the spiral, is constant. 
Many cylindric spiral springs in use have wire of square or of 
elliptic section. In another form which is used as a buffer 
spring, the diameter of the coils varies as if the wire had been 

wound on the half of a very 

bulging barrel-shaped mandril ; 

and we see that imder pressure 

this spiing can get very short 

/^ \ \ axiallv without the coils coming 

/ ^.-s 1 I into contact. In the form Fig. 

I ^^-^ / ^ 360, the mandril was of a coni- 

\ \ y cal shape. Lastly, in Fig. 361, 

X ^^ ^_ ^ we have the flat spiral spring — 

what the conical form would be- 

Fig. 361. come if it were squeezed until all 

its coils lay one plane, 

521. As an example of the bending of a strip of material, which 
might have heen considered after Art. 327, let ns take the case of 
a flat spiral spring, such as the main or halance spring in watches. 
Let xp M Fig-. 361) be such a spring, fastened to a case at >-. and 
to an arbor or axle at m. "S\Tien no forces are acting on the spring 
it has a spiral shape. Suppose that in this case, at a point p, the 
radius of curvature is r^, and that when the spring is partly wound 

up there is at p a radius of curvature r. then is the change 

r r^ 

. of curvature at p, and we know that the bending moment which 

prodtrced this change of curvature is equal to e i ( - - — ), "where 

E is the modulus of elasticity of the material and i is the moment 
of inertia of the cross-section. (Thus, taking e at 36,000,000, if 
the breadth of the spring is 0-2 inch and its thickness 0-03, then 
EI is 16-2.) Xow suppose the arbor to have turned through the 
angle x o g (which we shall call a) from the unstrained condition. 
What are the forces acting on p m and the arbor ? "Whatever 
these forces may be, they must be in equilibrium. If these forces 
were changed, there would be an alteration in the shape ; but so 
long as these forces do not change, the shape and position of things 
do not alter. This is why we can apply to the spring, p m, and the 
arbor the laws of forces acting on rigid bodies. So long as p m o 
does not alter in shape, it obeys the laws of rigid bodies. 

522. Xow, the forces acting on the arbor maybe very numerous 
—pressure of the pivots, pull of the f uzee chaiu, or pressure of teeth 
of wheels ; but whatever they may be, we know that they can be 
represented by one force acting at o, the centre, together with a 
couple, c. If the spring is not in contact with the top or bottom 
of its case, and if the coils are not in contact with one another, no 
other forces act on the spring, m p, except at p. The particles of 



APPLIED MECHANICS, 625 

steel on one side of the section at p are acting on the particles on 
the other side ; hut whatever the forces at each of the particles 
may he, we know that the total effect at p is the same as that of 
one force and one couple. We cannot easily say what the force is, 
hut if r is the radius of curvatui'e at p, and if r^ was the radius of 
curyature at p when the spring was unstrained, then the couple at 
p is what we have already called the hending moment, 

12 \t - ^J 

Let ns suppose, for simj^licity, that the spring is everywhere of 

the same hreadth and thickness, and let ns use the letter e instead 

E ht^ 
of -T^, which is now, of course, the same everywhere. The couple 

at p is then e{- V The only forces acting on p m o are : 

A force at o, of amount h, in the direction o h, say ; a couple at o 
whose moment is — l ; a force at p ; a conple at p whose moment 
is given aho^'e and is positive. Now, we know that the sum of the 
moments of all the forces ahout any point must he nothing. Take 
all the moments about the point p. The force at p has, then, no 
moment, and is to he neglected, and we have 

In fact, e\ )=rL + h. ph. 

Let p Q he a short distance measui-ed from p along the spring. 
Multiply every term of the ahove equation by p Q, and we find 

(PQ P Q\ 
|=L,PQ + H.PH.PCl. 

Now, when p a is very small it may he regarded as the arc of a 

P Q . 

circle whose radius is r ; consequently — ;- simply means the angle 

between the radius, or normal at p, and the normal at q ; in fact, 
it means the small angle which the tangent at p makes with the 

P Q, P Q, 

tangent at q. Thus — ;- simply means the change o^hich 



-HXPH— L + (? 



has occurred in the angle between the direction of the spring at p 
and the direction at a. If now, instead of considering what occm's 
at the point p, we take the point q, we shall get just a similar 
equation for another little length of the spring. Suppose we do 
this for every short length of the spring, and add up our results ; 

P Q, P Q 

we shall find that the sum of all terms snch as — — — means 

T r^ 

the change which has been produced in the angle, between the 
tangents to the spring, at its two ends. Thus, suppose the arbor 
has turned through the angle 0, and su]3pose that, whether or not 
the point of fastening at n has been moved, the direction of the 



626 APPLIED MECHANICS. 

spring at n has on the whole changed through an angle )8 ; then 
we find that the sum of all the above-mentioned teims amounts to 
— )8. {d may be called the amount of winding up of the sj^ring ; 
)8 may be called the amount of j'ielding in the fastening to the 
case.) Hence the sum of all the left-hand sides of all such equa- 
tions as the above is ^ (0 — j8). 

Now let us consider the right-hand sides of the equations. 
Evidently the sum of all such terms as l x pa will be l x length 
of spring ; say l I. The sum of all such terms asH x ph x pq 
is (Art. 109) equal to h multiplied by the length of the spring 
multiplied by the perpendicular distance of the centre of gravity 
of the spring from the line o h. This is, of coiu-se, the length of 
the spring multiplied by the moment of the force h about the 
centre of gravity of the spring. Summing up our results, we find 
that if the force on the arbor through the pivots, etc., has a 
moment about the centre of gTa^dty of the spring of the amoimt g, 
if the length of the spring is I, if the angle turned through by the 
arbor from the unstrained position is 6, and if /8 is the angular 
jdelding at n, and l is the couple with which the arbor tends to 

unwind itself, then e {e — fi) =^ hi — g I, or i. z=:~ {0 - /3) + g. 

The term g depends on the position of the centre of gravity of the 
spring. 

If the coDs are numerous, each will be nearly circular, and the 
centre of gravity of the spring will nearly be at o, and g becomes 

insignificant ; so that the equation becomes l =- (0 — )8). If the 

spring is so rigidly fastened at its ends that there is no change 

of direction relatively to the barrel, l =: -r 0, and the couple exerted 

by the spring in trying to unwind itself is simply proportional to 
the amoimt of turning of the arbor or the amount of winding up. 
If, then, the centre of gra^dty of the spring always remained in 
the centre of the arbor, and if the spring were rigidly fastened at 
N and M, we should have the couple exerted simply proportional 
to the angle of winding; and this is the condition for perfect 
isochronism in the balance S23ring of a watch. I need hardly say 
that this condition can never be j)erfectly satisfied. If we use a 
f uzee, the mainspring may be fastened as we please ; but sui)pose 
we want the couple exerted by the spring to be nearly constant for 
various amounts of winding up, it is e\ddent that the angle 3 
ought to increase as fast as Q ; that is, there ought to be a very 
considerable amoimt of yielding in the fastening of the spring to 
its case. The same effect will be produced by exerting consider- 
able pressure on the arbor at its pivots, or in some way causing 
the arbor and its case to be not quite concentric with one another. 
The watchmaker's usual plan to get moderately good iso- 
chronism is to make one of the above errors tend to correct 
another ; that is, by allowing a greater yielding or greater stiffness 



APPLIED MECHANICS 



627 



of the outer attachment to counteract the results due to centre of 
gravity of the spring not remaining exactly in the axis of the 
balance. 

523. Thus we see that by applying the law given in Art. 522 
to a flat spiral spring fastened to a case at its outer end, n, 
and to an arbor or axle at its inner end, m, we find that if the 
spring is riveted firmly both at n and m, and if it is so long and 
its coils so nearly circular that its centre of gravity is always 
nearly in the centre of the axle, then, when partly wound 
up, the spring tends to unwind itself with a turning moment 
which is proportional to the amount of winding up. This is 
the case in the balance spring, and it is this condition that 
gives to the balance its character of taking almost exactly the 
same time to make a small swing as to make a great one. 
(See Art. 455.) When the end n is not riveted, but merely 
hinged or fastened in any way that will allow it to turn about 
N, the unwinding tendency 
is not proportional to the 
amount of winding up ; it 
is proportional to the dif- 
ference between the angle 
of winding and this angu- 
lar yielding at n. If the 
strip is everywhere of the 
same breadth and thick- 
ness, the unwinding tendency is proportional to the moment of 
inertia of its own section — that is, to its breadth and to the 
cube of its thickness ; it is also proportional to the modulus of 
elasticity of the material used, and is inversel}^ proportional to 
the total length of the strip. Suppose we wind a cord round 
the barrel or case containing a mainspring of a watch whose 
arbor is fixed firmly, and, using a scale-pan with weights, we 
find the turning moment of the spring for various amounts of 
winding up. If we plot our results on squared paper, we shall 
find that the points lie in a curve like a o, bo, c o, or d o of 
Fig. 362, whereas for a balance spring we should get nearly a 
straight line through o. 

In Fig. 363 is represented an instrument which I have 
been in the habit of using in my laboratory,* to show the con- 
nection between the turning moment and the angular wind- 
ing in a fiat spiral spring. Difierent weights used at the end 
* The woodcutter has represented too large a weight and too thin a, spring. 




628 



APPLIED MECHANICS. 



of the string give different readings of the pointer. By means 
of such an apparatus we are enabled to verify the laws de- 
scribed above. When we have performed one set of experi- 
ments with a spring, another set may he made on the same 
spring with its length diminished or increased by means of 
the arrangement for clamping, shown in Fig. 321. In this 
way we can experiment with springs of different breadths and 
thicknesses, as w^ell as of different materials. 

524. The flat spiral spring just considered is a case of the bending 
of a strip of steel along its entire length. I will now take np a 
case in which there is no bending. Fig. 364 shows a cylindric 
spiral spring whose coils are very flat. Besides its own weight, it 




Fig. 363. 



is acted upon by two equal and opposite forces in tlie direction of 
its axis, the supporting force at n and a weight at m. Now let us 
consider the equilibrium of the portion of the sj^ring from any 
point p to M. Suppose the wire cut at p by a plane passing 
through the axis ; this section will be more and more nearly a 
cross-section normal to the axis of the wire, as the sph^als are 
more and more nearly horizontal. Let us regard it as a normal 



APPLIED MECHANICS. 



629 



isfbss-section of the wire. Now, whatever may he the stresses at 

this cross-section, they must balance all the other forces acting on 

p M — namely, the force f at m, which is axial, and the weight of 

p M, which is Yery nearly axial. If we neglect the weight of. p m, 

we have only to balance the force f acting 

at M. To do so we evidently need a shearing 

force, F, at p, distributed over the section, 

and a twisting torque which is equal to f . p h. 

It is easy to show that the shear is of much 

less importance than the torsion. Indeed, in 

many ways it is like the shearing force in 

beams (see Art. 369), and we shall neglect it. 

Again, since p h is the same for every part 

of the spring, every section of the wire is acted 

on by the same twisting couple, just as the 

shaft of Fig. 191 or the wire of Fig. 186 and 

its strength is calculated in the same way. 

Now, what is the amount of motion at m in 

consequence of this twist ? As the wire is 

everywhere twisted, just as if it were a straight 

wire fastened at one end whilst at the other 

end there were a force, f, acting at the end of 

an arm whose length is equal to p h, the radius 

of the coils of the spring, the amount of the 

motion of m is just the same as the motion of 

the end of such an arm attached to the straight 

wire. 

525. We have, then, the following pretty 
illustration (Fig, 365), which serves to keep 
the rule for spiral springs in our memory. 
Let two pieces of the same wire of the same 
length be taken ; one of them kept straight, 
fixed firmly at A, and fastened at b to the 
axis of a pulley which can move in roller 
bearings. A cord, c, fastened to the lim of 
this pulley, carries the upper end of a spiral 
spring, D E, formed of the other piece of wire, 
the diameter of its coils being equal to the 
diameter of the pulley. Evidently, if a weight, 
w, is placed in the scale-pan, a point e gets 
just double the motion of a point c, for B 
gets c's motion as well as the lengthening of 
the spring. The scales f and g and the Kttle 
pointers are for the purpose of making exact 
measurements. It is interesting to note how 
accurately the law is fulfilled, even in a roughly-constructed 
piece of apparatus such as anyone may easily put up for himself. 



^^^ 




Fig. 364. 



630 



APPLIED MECHANICS. 



Example. — A spiral spring of charcoal iron spring wire, 0*1 
inch diameter, 21-6 inches long, its coils having a radius of 
1*3 inch, is extended by a weight of 10 lbs. Supposing that 
a piece of wire of the same material 1 inch long and 0'05 inch 



-lil'' 




...:i-ilittifli Tig. S65- 

diameter, gets a twist of 2^-2 degrees with a 
twisting moment of 2 inch-pounds, what is the 
exten.sion of the spring] We see that if the 
trial wire were of twice the diameter the twist 
would be 24-2 -f- 16, or 1-51 degrees, and with 
a twisting moment of 13 inch-pounds, which 
is 6 '5 times as great, the angle would be 9-82 
degrees, and on a wire 21-6 times as long would 
be 212- degrees, or 3-7 radians, and the arc of 
a circle whose radius is 1*3 inch, subtending 
this angle is 3-7 x 1*3 or 4-8 inches, the 
answer. 

526. In designing a cylindric spiral spring it is 
very important to know the greatest elongation it will bear without 
taking a permanent set. If the material has internal strains given 
to it during its manufacture — and this it is yery difficult to prevent 
in steel springs, unless great care is taken in tempering, and it is 
almost impossible to prevent in brass spring-s, because the elasticity 
added in manufacture is often regarded as a necessary quality 
which ought not to be destroyed by any annealing process — in this 
case the reader must keep in mind the considerations of Art. 294a. 
Otherwise, let / be the greatest shearing stress per square inch 
which the material can resist without getting a permanent set. 
Let M be the greatest twisting moment which a round wire of 
diameter d can bear without getting a permanent set. We see 
from Art. 296 that 

ii = Trf3//16 .. .(1). 
Now / will be approximately known from Table XIY. , or m may be 
found by experiment for a given wire by any person who wishes 
to make a spring ; and whether ii or / is used in a formula, you 
now know how to calculate one when given the other and the size 
of the wire. If, then, we have a spring made of wire whose 



APPLIED MECHANICS. 631 

diameter is d, and if the radius of the coils as measured to the 
centre of the wire from the axis of the spring is r, we see that 
when w is the greatest weight with which the spring may be 
elongated without producing a permanent set, 

W = - = 7r^3//16 r (2), 

being independent of the length of wire employed. 

From Art. 295 we see that if n is the modulus of rigidity of the 
material, t' the greatest angular twist in radians which we can 
give to a wire of diameter d. inches and length 1 inch, and m! the 
twisting moment which produces this twist, then 

m' being what we have pre^dously measured or calculated, n is 
approximately known for a material fi'om Table XI Y., or t' may be 
found by experiment for a given wire ; and whether t' or n is 
used in a formula, you now know how to calculate one when given 
the other. 

Putting the result of our reasoning in Art. 524 into an algebraic 
form, we see that a load, w, wiU elongate the spring by the amount 

32 w^^ 
X = 5r .... (4), 

and hence the greatest elongation which can be given to the spring 
without its getting a permanent set is 

, , ^2lrm' 2lrf 
X z= Ir r, or -r, or — f .... (5). 

Combining (2) and (5), we see that when a spring is stretched 
to its elastic limit, the mechanical energy stored up in it, which is 
called its '< resilience,^'' being half the product of w', the proof load, 
into the proof elongation, is 

^ ,, irPdH 16Zm'2 

A mW A, or -^ — , or xr .... (6). 

527. Many interesting methods may be taken to express in words 
the meanings of these results. Thus the second expression in (6) 
shows that the work which we can store up in a spiral spring is 
simply proportional to the weight or quantity of material in it. It 
would be easy to show that we can store more energy in a spring 
formed of wire of cii'cular section than in one of equal weight of 
the same material whose wire has any other than a circular section. 

528. The following readings of our formulae may prove to be 
useful : — 1st. If d, the diameter of the wire, and r, the radius of 
the coils, be fixed, the elongation produced by any weight, w, will 
be proportional to I, the length coiled up to form the spring, 
2nd. If a wii^e of a certain length and diameter be given to form 
a spring, the elongation produced by a certain weight, w, wiU be 
proportional to the square of the diameter which we may adopt for the 
coil. 3rd. If the diameter of the wire be fixed, and the axial length 
of the spring, when closed, so that the coils may touch one another, 
or, what is the same, the number of coils be also fixed, I must be 
proportional to a, and therefore the elongation due to a weight, w, 



632 APPLIED MECHANICS. 

will be proportional to the third power of the radius which we may 
adopt "for the coU. -ith. If the length of the wii-e and the radius 
of the coil he fixed, the elongatioir due to a weight, w, will he 
inversely proportional to the fourth power of the diameter of the 
wire which we may adopt. 5th. With a given weight of metal 
and a given radius of the coil, the elongation due to a weight, w, 
will be proportional to P, or inversely to d^, since I must be 

proportional to -p. 

"SVe see that the ultimate elongation is — 1st, proportional to 
the length of the wii-e, if the diameter of the wire and the radius 
of the coil be fixed ; 2nd, proportional to the radius of the coil, 
if the length and the diameter of the wire be fixed ; 3rd. inversely 
proportional to the diameter of the wii-e, if the length of the wire 
and the radius of the coil be fixed. 

It will be foimd that a weight hung at m (Fig. 364) will tend 
to turn as the spring lengthens, imless the coils of the spring are 
very flat. This is due to the fact that the cross-sections of the wire 
- are really subjected to a little bending as well as torsion. 

529. We can cause the sti'ain in such a spring to consist altogether 
of bending, if, without exerting any axial force such as I have 
shown in Fig. 36i, we exert a couple about the axis such as we 
exerted on the wire in Fig. 186. The wire in Fig. 1S6 would be 
twisted, but the wire in Fig. 364 is subjected everywhei'e to bend- 
ing without any twisting, or with only a very little twisting, due 
to the fact that the coils are not perfectly flat. 

If rt'i^ is the radius of the coils to the centre line of the wire 

when unstrained, and the length of the coiled wire is /, then the 

nimiber of coils multiplied by the circumference of each is the total 

length, so that the number of coils is / 4- 2 iraQ. If now the 

moment of inertia of the cross-section of the wire about the axis 

through its centre about which it bends is i, and if m is the moment 

which acts at the unfixed end of the .spring to twist it, then the 

new radius, a, of every coil is obtained fi'om o ur knowledge of the 

fact given in Art. 325. 

M = E I X change of cm-vature, 

M 1 1 ,,, 

or __=___ (1), 

EI a fl'^, 
E being the modulus of elasticity of the material, i is bd^ -i- 12 
for a wire of rectangular section, d being the dimension in inches 
of the section, measured radially out from the axis of such a spring ; 
I is Trd^ -^ 64 for a wii-e of circular section of diameter d. 

Xow I -~ 'lira would be the number of windings ; so that, if 
n is the new nimiber and /?o the old number of windings, we have 

^.^ = "- ;<2'- 

But one additional windiug means 2 tt radians ; so that if 6 is the 
amount of winding up corresponding to 2 tt (« - Wo)> ^^ hiive 

. = ^.'..,.(3, 



APPLIED MECHANICS. 633 

We see that it does not depend upon the radius of the coils, and is, 
therefore, the same formula as given in Art. 522 for a flat spiral 
spring whose radius varied continually, e i is called the flexural 
rigidity of the wire, heing e bi^ [12 for a strip of thickness t ; being 
E Trd^fQi for a circular section of diameter d ; being e s^jli for a 
square of side s. The strength is as that of a beam subjected to 
the bending moment m. 

530. From these considerations it is evident that a spiral 
spring like Fig. 364, when it lengthens under the action of a 
weight, has all its wire subjected to torsion. The spring 
itself is extended, but the wire of the spring is twisted. 
Again, if we subject the spring to torsion as a whole, the 
strain really going on in the wire is a bending strain. 
Usually, a spiral spring, as its coils are not perfectly flat, has 
its wire subjected to torsion principally, and a little bending 
as well, when the spring is extended ; and when the spring is 
twisted as a whole its wire is mainly subjected to bending, but 
there is also a little twist in it. The extension of a spiral 
spring is proportional to the pulling force, and also to the 
length of the Avire and to the square of the diameter of the 
coils : it is inversely proportional to the fourth power of the 
diameter of the wire if the wire is round. The twist given to 
a spiral spring as a whole is proportional to the moment of the 
twisting forces — it does not depend on the size of the coils ; 
it is proportional to the length of wire, and inversely pro- 
portional to the fourth power of the diameter of the wire if 
the wire is round. 

531. We have taken up at length the two cases of spiral 
springs in wdiich the angle of spiral is 0. It will be found 
that v^hen the angle is not 0, the stiffness of the cylindric 
spiral spring follows much the same law as for springs of 
small angle, but it is necessary to take up the general case. 

532. The theory of the cylindric spiral spring is enough to 
study, because each small portion of any sj)iral sj)rir)g may be re- 
garded as part of a cylindxic spiral spring. A rough model will 
help a student to understand the work better. 

We shall imagine the upper end of a vertical cylindric spiral 
spring to be held fixed, and that at the other end, by means of a 
rigid arm coming in from the wii-e to the axis of the spring, we are 
able to apply an axial force f, by means of a weight, tending to 
elongate the spring, and also a couple, l, about the axis of the 
spii-al tending to increase the number of coils ; the directions of 
elongation axially and of greater winding up are our positive 
directions of motion. Let the axial elongation produced be called 
a?j and the angular rotation produced be called <p. "We shaH 



634 APPLIED MECHA>-ICa. 

neglect Hhe weight of the spring itself as it will "be quite easy 
afterwariis to correct for this. 

Let r be the ra<litL3 of the coils — that is, the distance of the 
centre of the wire everywhere from the axis. Let the whole length 
of wire "be /, the angle of the spiral a. Let b he the flexnral 
rigidity of the wire in the osculating plane of the spiral b = z i 
where e is Yoimg's mc^nlos of elasticity, and i is ^e moment oi 
inertia of the section of the wire about the line through its centre 
of gravity which touches the cvlindric surface, and is at right 
angles to osculating plane. The bending moment, divided by 
the flexural rigidity, gives the change of curvature produced 
by bending. Let x be the torsional rigidity of the wire. 
The twisting moment applied to a wire divided by x gives the 
angle of twist produced per unit length of wire. In Table XY. a 
num^ber of values of a and b are given for sections of wire which 
are in conmion use in springs, a is the torsional rigi'iity of the 
wire, being the twisting m.oment required to prc>duce unit angle of 
twist per unit length, b is the flexural rigidity of the wire in the 
osculating plane of the spiral, being the bending moment required 
to prcKiuce unit change of curvature in that plane. The line p q 
represents the axis of the spiral relatively to the wire. 

X is the modulus of rigidity, and e the Young's modulus for 
the material. In the last two cases t is supp<35ed to be small in 
C'l-mparison with b. Xotice that a iu the elliptic sections becomes 
STixPl^y when d is small iu comparison with d, and it becomes 
N^^/3 in the rectangular sections when f is small. Coulomb's 
wrong assumption was that a = x i when i is the moment of 
inertia of a section about its centre. Now, for the elliptic section 

I = — (d^ J -r D d^j , and hence 

Cc^nlomb's a ^ ly^d + J>d^ 



= i^^^(-.^;^i{-^r 



c-«:>rrect a 

if ar is d d. Coulomb's value is correct when the section is circular, 
and we see that it is more and more wrong as the section gets 
flatter and flatter. The true value of a is always easy to edcnlate 







APPLIED MECHANICS. 
TABLE XY. 



635 



A. 

Torsional 
rigiditj^ 





M 



^ 









32 



(D-* - d^) 



NTT d3(^3 



16 B- + d^ 



D^CP 



16 d2 + f^2 



0-14058 Ns^ 



3 iF'+T' 



3 i'^ + ^-^ 



B. 

Flexural 
rigidity. 



64 



64 



(d4 - ^*) 



64 



64 



T^^d 



12 



12 



E ^ 

12 



W. 
Axial load when /is 
greatest shear stress. 



TT ^3//16r 

(r = radius of coils) 



7r/(D4-#)/16D 



^■Dd^fjWr 



Dd'-fjUr 



s3/M-79r 



bHYIH'f^ + f^) 



Ptifl'd,r{b'^-Vt'^) 



636 APPLIED MECHANICS. 

in the case of an elliptic section. Its value for a rectangular 
section is calculated fi-om an infinite series, and it is therefore 
very important to know that Cauchy has proved that the torsional 
rigidity of a rectangle hears (approximately) to the torsional 
rigidity of an inscribed ellipse the proportion of their moments of 
inertia in the case when d is several times d (see Art. 313), 

Consider the portion of spring helow any cross-section p. This 
portion is in equilibrium. Hence the molecular forces exerted on 
It at the section p must balance f and l, and these are the only 
conditions -which we find it necessary to consider. Let Fig. 366 
represent the elevation of a portion of the wii-e beluw the section p. 
Let T T be the elevation of the axis of the spring. Then about the 
axis PM we know that the moment due to the force f must be 
balanced by a torque of molecular forces whose amount is f ;• ; and 
about the axis p t there must be a torque of molecular force of 
amount l. Eesolving these in the directions ps and pu in tne 
plane of the paper, which is tangential to the cylindric surface at p, 
we have about p s the torque 

F r sin. a - L cos. a, 
and about p u we have 

p ;• cos. a ^ I. sin. a. 

Xow the moment about p s means a bending moment which 
produces an angular change per unit length whose amount is 

Fr sin. a + L cos. o , 

B • • • • w ' 

and the moment about p u means a twisting moment which pro- 
duces an angle of twist per unit length of the amount 



Let us now consider how these two angular changes in unit 
length of the wire cause elongation and rotation at the bottom end 
of the spring. If the spring is sufficiently long, it is obvious that 
an axial elongation x and a rotation <^ will occur at the free end of 
the spring, for we can then imagine that the lateral motions due to 
all the elements of the wii-e exactly counteract one another. It is 
therefore only necessary for us to obtain fi'om the above expressions 
those elements which produce x and <^. Xow any rotation of the 
body below p about any axis can be resolved into equivalent 
rotations about other axes, according to the laws for the resolution 
of vectors generally. 

The rotation (1) about the axis ps is equivalent to (1) multiplied 
"by cos. a about pt, and to (1) multiplied by sin. a about pm. The 
rotation (2) about the axis p r is equivalent to (2) multiphed by 
sin. a about the axis p t, and to (2) multiplied by cos. a about p ii. 
Adding, we get, on the one hand, the rotation about p t, produced 
by the flexural and torsional strains in luiit length of the wire ; 
and this multiplied by /, the tr^tal length of the wire, gives ^. 
Adding, we get, on the other hand, the rotatipn about p m, produced 
by these same flexural and torsional strains ; and it is obvious that 
this rotation causes a point on the axis of any portion of the spring 



(b 

-y = p r sm. o cos 



APPLIED MECHANICS. 637 

below p to be lowered by a distance wbicb is obtained by multiply- 
ing the rotation by r, tbe radius of the spiral. Multiplying by I, 
we obtain the whole axial lengthening of the spring — that is, 
putting oiu- answer in its simplest shape, 

■°(a--b) + H~ + ^)---(^)- 

X /cos.-a sin.^aX . /I 1\ ... 

_=:f,.(^--- + ^ + LSin.aCos.a(--^)....(4), 

^ and X being the rotation and elongation jjrodiiced in a spiral spring 
by an axial force p and a couple l acting together. In the theory 
<!> and X are assumed to be very small. If we use 5 (p and S x for 
them, using (p for 2 tt « where n is the number of turns, and x for 
the axial length of the spring, so that x/n is what we sometimes 

call the pitch of the spii-al ; then — r- lirr = tan. 4) .... (5) 

and sin. (p = x/l . . . . (6), and by means of integration we can find 
accurately the effect of alteration in r and a as the spring changes 
in shape. Taking (3) and (4) as they stand, however, I find that 
young students can have no better easy mathematical exercises 
than are to be obtained by working problems on them. 

For example, let I, r, a, a and b for a spiral spring be given, we 
can calculate x and </> when given f and l, or if given x and (p we 
can calculate f and l. 

The very common case of a small, say a = 0, leads to 
(^ = l^/b . . . . (7) 
x^Tlr^JA. . . . (8). 
Thus if the spirals are very flat, whatever the nature of the section 
of the wire may be, the spring when subjected to axial force has no 
tendency to rotate at its end, and the s^Dring when subjected to a 
couple merely has no tendency to alter in axial length. 

Again, in any given spring, suppose we are informed that the 
rotation of the end is prevented, and that a force f acts, putting 
(p = we find l in terms of f from (3), and using this in (4) we 
have x in terms of f. Again, as in chronometer springs, if the 
axial elongation is prevented and only a torque l acts, put x — 
in (4) and find f in terms of l ; use this in (3) to find (p in terms of l. 

Let a = - or 45°. (3) and (4) become 

2*/'=-(i-i) + -(i + i)-...(9) 

2./fr = Pr(i+^i) + L(i-i)....(10). 

If we apply to (9) and (10) the first condition given above, that 
is, (p = 0, and an axial force f acting (9) gives l = - f r — - — 
80 that (10) gives 

. = ^....(11). 



^ 



638 



APPLIED MECHANICS. 



Again, if we apply to (9) and (10) the second condition giveu 

above, that is, x = and a torque l acting, (10) gives 

B - A 
F r = - L 



so that (9) gives 



<P 



B + A 



B + A 

. . . (12). 



In estimating for general purposes the effect of altering the 
angle a ; or the effect of constraint, such as preventing rotation of 
the end of a spring when an axial force is aj^plied ; or the effect of 
change of shscpe of section, etc., it is useful to remember that for 
most substances we may take it that apj)roximately e = 2-5n. It 
is very interesting to study the formulse (8), (10), (12), using the 
Table XV. for the A'alues of a and b for various forms of section. 
We shall leave tliis study to each student for himself — working out 
only the following interesting examples. 

Students will please work carefully the following fifteen 
exercises : — Take n = 10*2 x 10'', e = 25-5 x lO^. Apply an 
axial load of 1 lb. and find the elongation in each case. There are 
10 springs of wire, 100 inches long; diameter of coils 2 inches. The 
first has round wire of 0-1 inch diameter. All the others are of 
such sizes of section that there is the same total volume of material. 
Half the spirals ha^'e an angle of 45°. "When the exercises are 
finished, divide all the elongations by that of the first spring, and 
enter the results on such a table as the following : — 

TABLE XVI. — Relative Elongations eor Same Axial Loads. 





a = 


a = 


45° 




Whether the end 
isallowedfreedom 
to rotate or not. 


Eud free to 
rotate. 


End restrained 
from rotating. 


Solid round wu^e. 


1 


0-9 


0-889 


Hollow circular section, 
thickness of tube -jLth 
of outside diameter. 


0-196 


0-197 


0-195 


Square wire. 


1-132 


0-948 


0-912 


Eectangular wire, as in 
Fig. 6, Table XV. 
Breadth 4 times thick- 
ness. 


2-029 


1-108 


0-349 


Eectangular wire, as in 
Fig. 7. Breadth 4 times 
thickness. 


2-029 


2-544 


2-440 



APPLIED MECHANICS. 639 

In the theory of the spiral spring I considered the hend and 
twist given to each small portion of a spring, and assumed that the 
resultant action at the bottom of the spring was a rotation and an 
axial elongation — that is, that the lateral motions produced on the 
bottom by all angidar motions of portions of the spring exactly 
balanced one another. As a matter of fact, however, it is only when 
the spring is very long that these lateral actions balance. To take 
the lateral actions into account is of no practical importance in such 
si)rings as we ha-^^e been considering (cylindric spiral springs) ; but 
when the spires of a spring rapidly change their character, particu- 
larly in the case of Hat spiral springs, the lateral actions are of 
importaiice (see Art. 522). 

533. If any important object were to be served, I would calculate 
the relative strengths of all the springs under the various condi- 
tions here mentioned. It is only necessary to know the load i*", 
which, when producing a bending moment f r sin. a and a twisting 
moment rrcos.a in a wire of the j)articular section at the same 
time, shall be capable of just i)r educing permanent set. For a 
circular section this is well known — that is, it is Avell known that 
for a circular section the above two moments, acting together, are 
equivalent to a twisting moment 

¥r sin. a -f- ^/ y'^r-^ sin.^a + F'-^r'-^ cos.''^a. 
Ot 

Fr (1 + sin. a) 

if the material has equal strength to resist shearing and tensile 
stresses. This is merely (3) of Art, 298. We see, then, that 
for a round wire the breaking-load for a spring of angle a 
is less than for one of the same size of wire and length of wire 
and diameter of coils, but quite flat in its spii^als, in the ratio 
1 : (1 + sin. a.) Thus a sj)ring T\dth a = 45° has a strength only a 
little over one-half of a spring with flat coils. This is the reason 
why, when a weight has been hung from a spring which produces 
permanent set, the spring so very rapidly gets completely spoilt, 
even although there is a counteracting inlluence due to the coils 
becoming smaller in diameter. 

It is sufficient for most practical pm^iDoses, then, to say that if 
the load to produce permanent set in a romid wire spring is 1 when 
the coils are flat, the load which will jjroduce permanent set when 
the coils are not flat is 

1 -^ (1 -1- sin. a). 

Again, in springs with flat coils, all of radius 1", the loads to 
produce permanent set are simply equal to the twisting moments 
which, when applied to the wire, produce i)ermanent set; and 
hence the follovt-ing table is -s-ery useful. 

w, in the last colimin of Table XV., gives the load which will 
produce the maximum shear stress/ in each of the sections of wire 
in the case of cylindric spiral springs, the angle of the spii^al a 
being 0. Hence, if / is the proof shear stress of the material, w 
is the proof load. The student will calculate the numbers of 
Table XV. as an exercise 



640 



APPLIED MECHANICS. 



In the following table the load to j)roduce permanent set and 
the resilience are given for cylindiic spiral springs, so that 
students who are in the habit of only calculating the strength and 
stiffness of springs with round wire may be able to arrive easily at 
the strength and stiffness of springs made with other kinds of wire. 

Comparison of the loads which wiU produce permanent set in 
springs of the same diameter of coils and same area of cross-section 
of wire : in all of them a = — that is, the coils are supposed to 
be as flat as possible. This is really a comparison of the torsional 
strength moduli of sections of the same area. Let students 
calculate the various numbers here o-iven. 



TABLE XVII. 





Load to Produce 
Permanent Set. 


Resilience per Cubic Inch. 


Circular 


1 


1 


Hollow circular, thickness of \ 
tube -Jjj of outside diameter j 


2-733 


( More and more nearly 
( 2, as tube is thinner 
( and thinner. 


Square 


■828 


•938 


Eectangular wire, as in Fig. 6. ) 
Breadth, 4 times thickness ... \ 


•110 


•224 


Rectangular, as in Fig. 7. ) 
Breadth, 4 times thickness ... ) 


•110 


•224 



641 



CHAPTER XXIX. 

RESILIENCE OF SPRINGS. 

534 An examination of our formulae will show that for nearly 
round wire, even when a is considerable, the rotation ^ depends almost 
altogether on the couple l, and the elongation x depends almost 
altogether on the axial force f. But when the section of the wire 
is very different fi'om circular, the angle a enters very materially 
into the calculation. Many examples of great interest may be 
taken up. Nearly flat coils are presumed. 

1. If ?■ alters in a spring, and if we want the resilience of the 
material to be the same everywhere ; that is, the wire being round, 
if we want the material to be equally ready to break everywhere, 

we must make — constant. That is, if the coils get twice as large, 

the diameter of the wire becomes ^2, or 126 times a,s large. 

2. Eound wire all of same diameter d, but the coils capable of 
lying just within one another if the spring is compressed by axial 
force. 

Let H be the number of coils to any point starting from the end 
of the wire at the small coil side, where the radius is r^. We may 
take r = r„ + nd, I = Tq 2 irn + 4 Tr-u-d ; and adding together the 
elongations produced on each elementary length into which we may 
imagine the wire divided, wc find 

16f r \ 

The strength of this spring is to be calculated as if all the coils 
w^ere equal in size to the largest. 

Some students may perhaps be interested in working the 
problem : — Find the cylindric spiral spring which, made of the 
same wdre with the same number of turns, will receive the same 
axial elongation or compression from the same axial force. 

The answer is, if n is the total number of turns, e, is the radius 
of coils in the new" spring, 

U3 ^ r,S + 3 ,.2 ,^a ^ ,.^ (,,^)2 + 1 („^)3 

Thus, as a numerical examjDle, if the first and last coils were 4 and 
9 inches in radius, d = 1 inch, w r= 5 turns, b, is 6' 8 inches. 

535. Resilience. — The average resilience per cubic inch of 
a spring is the whole resilience divided bv the volume. It is 
only in the case of uniform stress and strain in the material 
everywhere that we have maximum resilience for the whole 
spring. In actual cases this only occurs in tie-rods and struts, 
and in spiral springs made of a wire which is a thiu tube. 



642 APPLIED MECHANICS. 

I M is the resilience per unit length of a wire, if M is the 
proof twisting moment and 6 is the proof angle of twist, or il 
M is the proof bending moment and 6 is the proof change of 
curvature produced. It is easy, therefore, to work out the 
following table of values, and every student ought to do this 
as an exercise : — 

TABLE XVITI. — Eesiliexce per Unit Volume in Inch-pounds. 

Simple compression or extension. Only convenient in) ^/- .r.^^ 

springs made of indianibber ) 2 — 

Bending if the maximum stress is reached in every ^t 

section, and section is rectangular. As in beams of i /-o 
miiform stxength, in u- ell-made carriage springs, and )> g-— = 133 

in spiral springs subjected to a torque only ; also | ^ 

C-springs. Art. 370 J 

Bendiag. Uniform strip fastened at one end and loaded j ^2 

at the other, or supported at the ends and loaded in > j^j-— = 44 

middle; ... .".'• ' ..'. ) ^ 

Simple shearing. Possible in springs of indiarubber ; ^ ^-o 

also possible in spiral springs made of thin tubes / V— =1,000 

circular in section ... ... ... ) ^ 

Torsion. As in spiral springs of round wire subjected to | j^/- -^q 

axial loads / - ^ 

Torsion. As in spiral springs of square wire subjected) .00- f~ 4^9 

to axial forces. j "' ' ^ 

Torsion. As in spiral springs of oval or rectangular wire subjected 
to axial forces, anything less than for circular depending on ratio of 
diameters or of breadth to thickness. 

536. The student may not yet have been sufficiently im- 
pressed with the importance of knowing the resilience per unit 
volume of the material in all the springs. 

The resilience per unit volume tells us the value of a 
particular shape of spring. 

Suppose that a spiral spring of any given material is to be 
used for any purpose. 

The greatest load is stated, and the elongation or com- 
pression due to that load. Theu whate\er other conditions 
may be given as to the coils being of the different sizes, 
etc., we know that the spring may be made of any shaped 
section of wire, but that the thin' tubular circular section 
is the very best and the solid circular section is half as 
good, and any other sections are not half as good, and that the 
spiral spring form is better than any other. 

In any spring, if a force f produces a known motion x in 



APPLIED MECHANICS. 643 

its own direction, then the resilience per unit volume, multi- 
plied by the volume of material, gives the proof load f' multi- 
plied by half the motion x produced by it. This principle 
enables easy calculations to be made. 

Again, if in any spring a torque L produces a known 
angular motion in its own direction, then the resilience per 
unit volume, multiplied by the volume of material, gives the 
proof load l' multiplied by half the angular motion 0' produced 
by it. 

In nearly any case we consider, it will be found that the 
important thing looked for when we choose a particular shape 
of spring is total resilience — total energy stored up. 

Thus for example : We want a spring to exert a force f' 
and only to alter, say, lo lb., for a change of shape indicated by 
a motion in the direction of f' by, say, h inches. Evidently here 
the law of stiffness of the spring is 6 aw, say h :=kw, where 
^ is a known number. 

And, therefore, if x is the greatest motion, x = - f' : and 

the total resilience is 77 f' x, or 7, — f'^. 

That is, we are given f', w^ and 6, and therefore the total 
resilience of the spring, and if we know the type of the spring 
we can find the volume of the substance required and therefore 
its weight. What people generally mean by the "springi- 
ness " of a spring means that it shall not change much in the 
force with which it acts, for a considerable amount of alteration 

in shape. Now h is the change of shape and — r is the fractional 

F 

change of shape, so that — f' represents the spri7iginess of a 

spring. But the value of a spring also depends upon the 

greatest force it can exert. That is, its value depends on — f'=^; 

that is, on its resilience. 

Looked at from almost any point of view, we see that the 
value of a particular form of spring is represented mainly by 
its resilience. Of course, its shape and cost of manufacture are 
also of importance. We may know that a tubular spiral spring- 
may be the best for weight, and yet a C-spring shape may be 
best suited to the use to which it is to be put, and we put up 



644 APPLIED MECHANICS. 

with the disadvantage of using from twenty-two to twenty-eight 
times the weight of metal because of some other convenience. 

537. If the angle of the spiral is o, the length of each turn, instead 
of being 2v r, as we take it in approximate calculation^, is really 
2 irri sin. a. If the axial length of a spring is x, the angle o is 
such that cos. a = x //. If the axial length changes to x -f :r, the 
angle changes to o^, such that cos. a^ — (x -f x]/l. TTe have not 
taken these matters into account, assuming that our elongation." 
were small or that our calculations were to be only approximate. 
The student who knows a little calculus may use 5 x and 5 (p instead 
of X and <p in Art. 532, and by integration obtiiin general and 
accurate expressions. 

A conical spiral spring of round wire of diameter d, whose 
spires vary gradually from greatest radius r^ to smallest ?>, ; the 
proof load is evidently to be calculated from t\ or w^ = T^f*/ 16>-i. 
The elongation for a given load w ought to be calculated for short 
lengths and added up. ^lathematicaUy it is evident that 

" w 



Taking it that if -s is distance from, the end of the wire where r = 
r = r^ + a ^— ^^ — ^, and it is easy to show that 
32 w I , , 



In fact, we see that in a conical spring, instead of r^ for a cylindrio 

sprin^r we take 

1 - o 



EXERCISES. 

1. If a conical spring is formed of round wire 0*2 inch diameter, 40 
inches long, the coils varying from r^ = 2 inches to r,-, = 1 inch, what is 
the proof load, and the axial shortening with this load ? 

Ans., 47 lbs. if/= 60,000 ; 0-4 inch. 

2. An AyTLon-Perry spring is made of strip section, as in Table XY., 
which nearly covers a cylmdric surface, and a = 45"^ so that ^^ = 2 xr x if 
X is the axial length, and x = ? sin. a, or 1/ v^. Hence lb = 2 irrl/ \^, 
b = -K V^r. If, then, r = 0-1 inch, b — -44 inch. If ^ = -001 inch, and 
we have been able to obtain very broad strips of steel of this thinness, 
A — -0019 B r= -00132 according to" Table XV. Hence a load F will pro- 
duce an elongation x = 128 F/ and a rotation ^ = — 11 "o F/. It is to be 
noticed that the strip section of wire in a spiral spring (as, for example, 
in what are sometimes called volute springs for buffers) is very wasteful 
of metal for ordinary purposes. 

3. Spring of strip 4 inches parallel to the axis 0-25 inch thick. If a 
is taken to be 0, and the coils lie inside one another, touching ; if there 
are n turns, the smallest of radius r^ and the largest of radius rj, the 



APPLIED MECHANICS. 645 

thickness of the strip being t, then the average radius is ^ (rj + r^,), the 

length is ? = 7r(rx + r^Jn and nt = )\ - r^, so that I = -7 (*i^ - ^'o^)- 

Hence, as when t is small compared with b, w of column 3, Table XV., 
becomes hfifjZr, we have in this case 

wi = le-fizn .... (1). 

(8) of Art. 532 gives for the shortening under a load s, x =■ Tlr^/~bt^, 

o 

and as in the case of our conical spring we must take instead of the 
constant r^ the value | {t\^ + ViVq + r^^), we have 



^{ri'-ro')Ur^^ + r^r, + r,'^)/~bt^ 



F TT 



^ = zrjH O'l + ''0) (^'1^ - »o')- 



EXEECISES ON SPEINGS. 

Take proof /= 140,000 lbs. per square inch, n = 13 x 10' lbs. per 
square inch for the best spring steel. 

1. A spring of coils 4 inches in radius is of round steel, 1 inch in 
diameter and 12 feet long. What are its proof axial load, its proof 
shortening or lengthening, and its resilience ? 

2. A spring of round steel wire the radius of whose coils is 2 inches 
is to be 20 inches long when its coils lie close together ; it is to elongate 
2 inches for a load of 400 lbs., and this is to be its proof load. Give its 
dimensions. 

Here, if n is the niunber of coils, n 2Trr = I nearly, nd = 20, so that 

1/2 irr = 20/f/ ' (1), 

400 = ird^ X 14 X lO^/lGr (2), 

2 = 32 X 400/?-2/irN(?^ (3). 

We have only to £nd I, r, d from equations (1), (2), and (3). 

3. A safety-valve spring of square steel wire is to shorten 0-4 inch 
when the pressure of steam is 150 lbs. per square inch, the effective area 
of the valve then being 12 square inches. The radius of the coils is 
2 inches. A load of 400 lbs. per square inch on the valve produces the 
proof load on the spring. Find / and c?. 



640 



CHAPTER XXX. 

CARRIAGE SPRIVGS. 



538. Carriage springs usually consist of strips of steel in 
contacr, as sLo^^ti in Fig. 367. They are fastened at the ends 
and middle to the objects through which the loads are applied, 
in various waySj which must be examined by the student in 
actual examples. The ends of the strips are usually shaped, 
as in Fig. 36S or Fig. 369. These plates, before being 
tempered, are curved very accurately to a template ; whilst 
dark red at the end of this shaping they are dipped into cold 
water, the two ends entering the water first and then the whole 
plate being lowered beneath the suilace, so that if there is a 
variation in the hardness at different places it shall be a 
symmetrical variation. Each plate has a little pin or feather 
which enters a slot in its neighbour, so that the plates shall not 
be laterally displaced relatively to one another. Sometimes 
this is effected by making a short wrinkle or corrugation length- 
wise at the end of each plate. A bolt passes through holes in 
the middles of all the plates, and when its nut is tightened up 
the plates are made to lie closely against one another. When 
springs of the general shape of Fig. 367, or of half of it, are 
made of one solid piece instead of a number of plates, the 
calculation of strenirth and stiffness is made bv the method of 
Art 339. 

539. Carriage springs are usually t€St€d by means of a 
hydraulic press or a steam p»ress which forces them to become 
quite straight three or more times, and a spring is thought to be 
satisfactoiily made if after this it is found not to have taken 
any permanent set. When the stndent works the following 
exercises, he will see that if the greatest load is that which pro- 
duces straightness in all the strips, their initial curvatures 
ought to be the same. This will be found also to give 
sufficient tightness when the plates are bolted up. 

I think that carriage spiing makers and buyers are too 
particular in their wisliing to have all the plates lying very 
rightly together when bolted up. There is too much of such 
tightness. TTsually. too, in actual use a carriage spring never 
b^x>mes unloaded, and its plates are therefoi-e always very 



APPLIED MECHANICS. 647 

much more tightly pressing together than when the spring is 
examined unloaded. If, however, such tightness is really 
thought to be necessary, and if to obtain it we must have such 
great differences in curvature as I have found in springs by 
the best makers, it would be advisable to make the shorter 
plates thinner than the rest. 

But if all the plates are of the same thickness, they ought 
certainly to have the same curvature before bolting up. If the 
student draws a set of such plates, he will see that when they 
are bolted up they will be sufficiently tight, and if such a 
spring is straightened the stresses in all the plates will be 
exactly the same. 

540. The following exercises on bending are to be worked 
by the student to lead him to the theory of these springs. 
These exercises deal almost altogether with the best construc- 
tion of carriage springs. 

EXERCISES. 

1. A beam of constant strength ; what is its curvature everywhere ? 
Ans., the strength modulus z is i -^ ^d i£ dis the depth at any place 

and I is the moment of inertia of the cross-section there. Hence --d = f, 

2 1 

M 2f 
a constant for all sections, heing the o-reatest stress in all, or — == — : 

El -^^Z 
and this is the curvature. 

2. If a beam has had the same change of curvatiu-e everywhere, show- 
that if it is of constant strength it must be of constant depth. 

3. If a beam of rectang-ular section of constant breadth and variable 
depth d is loaded at one end and fixed at the other, what must its depth 
everywhere be for the cm-vature to be constant ? 

Ans., if X is distance fi'om the load w, ^ must be constant, and 

therefore dec %/ x. 

4. In the last example, what is the greatest stress in each section ? 

—r-j2 =/> so that fee ^, or j <x —^, or cc x^. In fact, / is simply 

proportional to the depth, if the depth is proportional to the cube root 
of X. If we compare the answers to exercises (2) and (4), we see that the 
overlapping parts in caiTiage springs are most economically made of 
constant depth but varying- breadth. 

5. n strips, each of thickness t, make up a carriage spring if the 
length of the top strip (or the length of the curve a c b) is 2 ? inches ; the 
overlap being x, I = nX. If o c is small compared with r^, the radius of 
curvature of the top strip, = 1-/2 ri nearly. Hence change in o c = 
^ /2 X change of cm-vature of top strip. 



648 



APPLIED MECHANICS. 



6. If when loaded the radius of the top strip is Bj, instead of "being 
infinite, what is the radius of every strip when the spring is taken apart 

if the resilience is to be 
A 2: -^^ uniform? Also what 



is i\. the radius of the 
top strip in the un- 
loaded spring ? This 
is a more tedious 
exercise, but is easy 
enough. 
Am. Fii'&t, if the radius of the top stiip when separate is pi, pi la 

known, because = Ifi e i, and f is the maximum stress. 




1 _ \^ 
^\ pi 



Ki + St 



(1) 



where p^ is the radius of the s^" strip when separate. We no-w 
have the interesting problem : When a number of strips whose 
radii when free are known, are fastened together to form a spring, 
what will ^'1 now be? The radius of the s'^ strip was p^ when 

free, and is now ri + st, and so its change of curvature is 
I I 

-. Hence the force w, at its end, producins; the 

Ps n + st , F ^ 

w^ 
bending moment ws A, must be such that produces this change 

of curvature. 



Hence 

EI / 1 
W^, = — ( — 



n + st 



) . . . . (2). 



We must, therefore, write out equation (1) for every strip to find 
the values of p^ for each, in terms of pi, which is known, and also 
equation (2), thus calculating every value of w^. the imknown r^ 
being in. every term. Now we make the statement that the sum of 
all the values of w^ is 0, and this enables ri to be foxmd. Thus (2) 
and (1) give 

EI/ 1 11 1 \ .... 

^ A \Ri + SC Ri pi >i + si) ^ " 

and we must find i\ by inserting 1, 2, 3, etc., for s in the equation 

If st is small compared with I'l and ki, we find that, if there are n 
strips altogether, 

' ^-^ ' ^ ^ ■ /., _ 1 \ I ,-,2 _r, + i f fn - 1) = 



■Et 



ai 



i?l^^' 



1) 



(5), 



a quadratic to find r^. 

It is not necessary to pursue the subject farther. Student* 
have the means of working all sorts of practical problems. 



APPLIED MECHANICS. 649 

541. Tempering Carriage Springs. — In the last part of the 
work on the strips, a plate being red hot, by an interesting 
manipulation between several pairs of smiths' tongs held by two 
workmen it is fitted to lie everywhere close to a curved piece of 
metal, so that it receives a definite shape. It is now a dark 
red, and the workman dips the plate into w^ater, letting the 
two ends go in simultaneously and the middle of the plate last. 
The plate is now placed in an air furnace, where it gradually 
heats ; it is taken out once or twice and rubbed with a piece of 
partly charred wood. The experienced workman can tell by 
the nature of the smoke coming from the wood whether the 
temperature of the plate is high enough ; when it has been 
heated to a sufficiently high temperature it is withdrawn from 
the air furnace and allowed to cool, lying on a metal table with 
others which have preceded it. 

542. Tempering Spiral Springs. — The processes employed by 
Messrs. Salter are, unfortunately, unknown to me. In all cases, 
however, they probably consist in winding wire or rod in the 
red hot state on a smooth iron mandrel, shaping the end parts 
of the spring, if the wire is small, by pliers ; if large, by special 
tools which can readily be designed for special cases. The 
formed spring is now heated to a dull red heat in an air 
furnace and plunged into hot oil, just of such a temperature as 
will produce the blue colour peculiar to spring steel. It is in 
this dipping of the hot steel into the oil bath that any trade 
secret can exist. If all the steel could from the same instant 
and at the same rate lose its heat, so that the hardened steel 
would be uniformly hard everywhere, the process would be 
perfect. Large spiral springs for safety-valves are dipped so 
that their axes remain vertical. To lay them sideways into 
the bath might cause the lower half of each coil to cool more 
quickly than the upper half. It is quite possible that good 
spring makers may not only dip their springs axially but also 
give to them a rotatory motion round their axes as they enter 
the bath. 

543. Very small spiral springs for the balances of chrono- 
meters and watches require special care, whether they are cylin- 
dric or flat spiral springs. This is on account of the thinness 
of the material and the rapidity with which it may cool. They 
are usually enclosed in a box — very small springs are enclosed 
in a platinum box — either on their mandrels if cylindric, or 
coiled up together with others if flat, surrounded by powdered 



650 APPLIED MECHANICS. 

charcoal. The box is heated to redness for a sufficient time to 
ensure the springs inside being all red hot, and the box is then 
immersed in water. The little box is manipulated by means of 
a long handle. It is now placed upon an iron pan placed over 
a flame, and lying beside the box on the pan is a piece of 
brightened steel. As the pan gets heated the brightened 
piece of steel takes different colours, and it is presumed that 
the steel springs have about the same temperature. When the 
dark blue colour is reached, which is so characteristic of spring 
steel, the box is removed from the pan and allowed to cool. 

544. Tempering Flat Spiral Springs.— In many cases these 
springs are not shaped before tempering. Great lengths of 
steel strip are passed through an air furnace from one 
set of rollers to another, at a rate which depends upon the 
particular purpose for which the strip is designed, upon its 
breadth and thickness, upon whether it is Bessemer or crucible 
steel. Just when in the red hot state and leaving the air 
furnace, the strip is passed through a vessel through which 
cold water is kept circulating, and is consequently made very 
hard. It then passes between pieces of cotton waste kept well 
soaked in oil, and then over a flame which keeps the oil ignited ; 
and on leaving this region the strip is gradually allowed to 
cool before being wound upon a roller. If allowed to move too 
slowly, the strip and the oil covering its surface are for too 
long a time subjected to the heating effects of the second 
furnace, and the steel becomes softer. I have seen strips which 
were said to be continuous and a mile long which had been 
tempered in this way. 

The very fine strip steel which is used in the Waterbury 
watch is tempered in this way. Great quantities of cheap 
Bessemer steel strip for use in ladies' corsets and numerous 
other purposes are also tempered in this way. 

545. I have never seen this process carried out by electrical 
heating as a substitute for the air furnace, but it is obviously 
quite easy to make the substitution. The strip or wire to be 
tempered receives an electrical current from a certain roller A, 
over which it moves, and another roller b at a short distance 
from the first ; and at c, or even before it reaches B, the steel 
passes into an oil bath. In this case we have no sudden great 
hardening and then a softening process ; the tempering is all 
done in one operation, a cooling from red heat to the tempera- 
ture of hot oil As tlie wire or strip is still kept heated when 



APPLIED MECHANICS. 651 

in the oil, the cooling is gradual, and obviously the softness of 
the steel will depend on how much of it is kept heated in the 
oil and on the average temperature of the oil. 

546. In well-made carriage springs the resilience is 

1 f^ 

^ — per cubic inch, 

and in designing a spring a knowledge of this fact enables us 
somewhat to shorten the work. 

Thus, suppose a spring is wanted to take a proof load of 
3 tons, with a deflection of 3 inches just making it straight. 
The total resilience is 

I X 3 X 2,240 X 3, or 10,080 inch pounds. 

If/' = 30,000 and e = 30,000,000, the resilience per cubic 
inch may be 5 inch pounds, so that the volume of steel required 
for the spring is 2,016 cubic inches. Now if there are opiates, 
this volume may be taken as 

nhtl {I being half the length of the longest strip), 
or nbtl = 2,016 .... (1), 
which is one equation towards the solution of some problem. 

EXERCISES. 

1. What ought the initial curvature of a ^-inch plate to be if the 
proof stress of the material is 30,000 lbs. per square inch (exists when tho 
plate is straight) and e is 30,000,000 ? 

Here, by formula of Art. 540, 

30,000= 15,000,000 x ^w, 
w= -004. 

The radius of curvature is -r^ = 250 inches. 

2. If the above plate is 36 inches long, what is its initial dip^ oe, in 
Fig-. 367 ? Using the approximate formula oe = l^ -i- 2 r, we have 

3. If the spring is fonned of six plates, the longest heing 36 inches, 
the overlap on one side of plate on its neighbour is 

36 -^ 12, or 3 inches. 

4. If the breadth of each plate is 3 inches, 

u =3 -000259 w. 

5. The load w' which will produce the deflection 0*64 inch — that is, 
which will straighten the spring — is 



652 APPLIED MECHANICS. 

6. How many plates of f inch thick and 3 inches broad, the longest 
being 30 inches in length, of the above-mentioned steel will be required 
for a spring whose proof load is to be 1 ton ? 

_ 2 » X 3 X -i^ X 30,000 
' 3 X 30 ' 

or n = 8. 
Hence the spring ought to be made of eight plates. 

547. The answers to the above exercises guide us as to the 
best way of constructing carriage springs so that the strips 
throughout shall have the strength of the overlapping parts. 
The following rules are based upon the dimensions of these 
overlapping parts, v.^iich are merely little cantilevers. 

If X is the overlap, Fig. 367, and I is the half-length of the 
longest strip, so that oi being the number of strips 7i\ =z I , 
then t being thickness of each strip, b its breadth, d the deflec- 
tion of the spring, /the maximum stress, E Young's modulus, 
w the load at a and at B, there being a load 2 w at the middle, 
it is evident that 

/= 6wl ^ nbt%D = 6wl^/nBbtK 

If/' is the proof stress, the resilience in a well-made spring 
is/'^ -f- 6 E inch pounds. Carriage springs are usually made of 
Bessemer steel, and we may take /' = 30,000 lbs. per square 
inch and e = 30,000,000. 

Examples. — 1. If the overlap \ is 2 inches, and there are 

10 strips each f inch thick, so that the half-length of the whole 

spring is ^ = 20 inches, ^ = f , and if 6 = 2| inches ; find the 

load which wdll deflect the spring 2 inches. Here d = 2 

2 = 6 w 203 -^ 10 X 3 X 107 X 21 X (|)3^ and hence 

w = 5-926 X 105 lbs. 

If the plates were not of the same initial curvature — that 
is, if the spring was, in this respect only, badly made, then the 
law for deflection is still true for the badly made spring ; but 
the rules for strength are untrue. We only calculate the 
increase in / due to a load w. 

548. In the above theory I have assumed no friction between 
the plates. This friction makes the bending to be less for a given 
load if the load has been increasing, but if the load has been 
diminishing the bending will be greater than the formula (1) 
gives, on account of friction. When we test a carriage spring, 
the effect of friction in causing the deflections to be greater with 
diminishing than with increasing loads is very noticeable. I 
have already pointed out the unsuitableness (on this account) 



APPLIED MECHANICS. 



653 



of these springs for all measuring purposes, and how suitable 
they are for carriages and under other conditions where vibra- 
tions must be rapidly stilled by frictional forces. 

549. In a spring such as Fig. 367, if 2 w is the upward load 




Pig. 368. 



Fi-. 



applied at the middle, the downward supporting forces at a and 
B are each w, and the end of each plate may be regarded as 
applying to the plate above it this force w. Thus in Fig. 368, 
if A B is the overlap of one plate on another, every part of the 
plate except the overlap part ab at each end is subjected to a 
bending moment w . A b everywhere, so that the change of 
curvature everywhere is the same. 

Also the part A b oughb to be fashioned like a beam of 
uniform strength and curvature fixed at b and loaded at A. 
We find from Exercise 2, Art. 540, that we can make it of 




Fig. 370. 



Fig. 371. 



uniform strength and uniform curvature at the same time by 
varying the breadth of the plate but not its thickness, the 
proper shape being shown in Fig. 370. This is roughly 
approximated to in many springs where the ends are shaped 
as in Fig. 368. 

If we keep the breadth constant, as in Fig. 369, but alter 
the thickness so as to have constant curvature everywhere 
(so that the end part of one plate may fit properly against 
another plate), we see from Exercises 3 and 4 that the thick- 
ness ought to vary as the cube root of the distance from a ; 
but it is not possible to have also uniform strength in this 
part of the plate. 



APPENDIX. 



TABLE XIX.— Useful Constants. 



Time. 



One sidereal day 
INlean solar day- 
One year 



86,164 seconds. 
86,400 seconds. 
36524224 mean solar days. 



Length. 



British standard; tlie yard = 3 feet = 36 inches. 

1 chain z= 66 feet =100 links = 4 perches. 

1 mile = 1,760 yards = 5,280 feet = 80 chains = 8 furlonj^ 

1 nautical mile := 6,080 feet (average). 

zzzlO cables zi: 1,000 fathoms (nautical). 
Telegraph poles 220 feet apart. 
1 fathom = 6 feet. 



3-2809 feet. 



Surface. 



French standard, the metre = 39-37 inches : 

1 kilometre = -6214 miles. 
1 inch = -0254 metre = 2-54 centimetres. 
1 foot = 30-48 centimetres. 
1 yard = -9144 metres. 
1 mile == 1,609 metres = 1-609 kilometres. 



1 square inch = 6*451 square centimetres. 

1 acre iz: 4,8 JO square yards. 

1 square mile rz: 640 acres z= 2-o9 square kilometres. 



Volume. 1 cubic inch = 16-387 cubic centimetres. 

1 cubic foot = -02832 cubic metre = 28-31 litres. 
1 litre = -22 gallon. 

For many practical calculations it is sufficiently correct to take : — 
Length. 1 inch = 25 millimetres, but 25*4 is more correct. 

1 metre == 3 feet 3 inches and f ths of an inch. 

10 metres =11 yards. 
20 metres = 1 chain. 
8 kilometres = 5 miles. 

Surface. 1 square inch = 6^ square centimetres. 
6 square yards = 5 square metres. 
1 acre = 4,000 square metres. 

Volume. 4 cubic yards = 3 cubic metres. 
1 gallon = 4| litres. 

4 litres = 7 pints. 

100 litres = 22 gallons. 
1 cubic foot =28-3 litres. 

Weight. 1 gramme = 15| grains. 

10 kilogrammes = 22 pounds. 
50 kilogrammes = 1 hundredweight. 
1,000 lologrammes = 1 English ton. 

Speed. 1 knot=l nautical mile per hour = 1-15 miles per hour= 1*7 
ft. per sec. =101 ft. per min. =51*5 centimetres per sec 



APPLIED MECHANICS. 655 

60 noiles per hour=: SS feet per second. 
Weight and Force. Britisli engineer's unit of force =the weight of the 
standard poimd in London. 
Weight of 1 lb. = 16 czs z= 7,000 grains = 453-6 grammes = 
445,000 dynes = -4536 kilogrammes. 1 oz. = 28-35 grammes. 
"Weight of 1 grain = 63-57 dynes. 
I kilogramme = 2-2046 lbs. 

1 ton = 2,240 lbs. = 1,016 kilogrammes. 

1,000 kilogrammes = 1 ton (metric) = 2,205 lbs. = -9842 ton (British). 

^ = 981 centimetres per second per second. 
= 32-2 feet „ „ 

Value of ff at London = 32-182 feet per second per second. 
„ „ Equator = 32-088 „ „ 

,, ,, the Poles == c;2-253 ,, ,, „ 

Inertia or mass of a body = weight in lbs. at London -~ 32-2. 

1 gallon of water at 62° F. weighs 10 lbs. by English law. 
1 cubic foot of water weighs 62-3 lbs. 

1 cubic foot of air at 0° C. and 1 atmosphere weighs -0807 lb. 
1 cubic foot of hydrogen „ ,, „ ,, -00557 1b. 

For academic calculations we usually take the weights in pounds of 
1 cubic foot of each of the following to be : — 
Brickwork, 112; concrete, 150; grindstone and Portland stone, 
131 ; granite and marble, 164 ; dry oak, 58 ; dry fir, 47 ; cork, 
15; coal, 80; clay, 120. 
Weights per cubic inch of each of the following materials, in 
pounds — cast iron, -26 ; wi-ought iron, -28 ; steel, '28 ; brass, '3 ; 
copper, -32; bronze, -3; lead, 4; tin, -27; zinc, '26. 

Work, Energy. 1 erg = 1 dyne x 1 centimetre. 
1 gramme- centimetre = 981 ergs. 
1 foot pound = 1 -356 X 10^ ergs. 

1 kilogramme metre = 7*233 foot pounds. 
1 horse-power = 33,000 ft. lbs. per min. = 550 ft. lbs. per 

sec. = 7 *46 X lO'* ergs per sec. = 746 watts (watts == 

volts X amperes). 
1 kilo-watt = 1,000 watts = 1*34 horse power. 
1 force-de-cheval = -9863 horse-power. 

Energy obtainable from 1 lb. of coal = 8,500 Centigrade heat 
imits = 15,000 Fahrenheit heat units =: 12 x 10^ foot x>ounds. 
Joule's equivalent — 

1 pound Fahrenheit imit of heat = 780 foot pounds. 
1 pound Centigrade unit of heat= 1,400 foot pounds. 
See note, p. 42. 

1 horse-power horn- = 1-98 x 10^ = 2 x 10^ ft. -lbs. nearly. 
1 Board of Trade electric imit= 1,000 watts for 1 hour = 

-iTT- r^ horse-power hour =z 2-654 x 10* ft.-lbs. 
/46 

The base of the Napierian logarithms is e = 2-7 1 8. To convert com- 
mon logarithms into Napierian logarithms, multiply by 2303. 



656 



APPLIED MECHANICS, 



TABLE XX 


— AIODUL 


[ OF Rig 


DITT. 






y = M Villus of Rigidity in Malion? 






: f Pounds per Square Inch. 


Steel Plates, ^ per cent. 


Carbon -^ 






»» » 2 » 


;; i - 




13 


J» » 1 »> 






Steel BoUer Plates 


... 




13o 


Cast Steel (tempered) .. 








U-0 


„ „ (untempered) 








120 










110 


„ „ (hardened' 








110 


Cast Iron 








50 to 7-6 


Iron Boiler Plates 








U-0 


Wrouglit Iron B^ts 








lO-o 


„ Plates .. 








9-0 


Soft Iron 








108 to 11 


Brass 








5 to 0-0 


Copper 








5-6 to 6-7 


Lead 








•27 


Zinc 








5-1 to 0-4 


Tin 








2-2 


Gold 








4-0 to 5-6 


Silver 








3-8 


Platinum 








8-9 to 9-4 


Aluminium 








3-4 to 4-8 


Delta Metal (rolled) . 








5-25 


G«n „ 








3-7 


ITiosphor Bronze 








5-2 


Glas* 








3-3 to 3-9 


Wood 








•1 to -17 


Granite 








1-8 


Marble 








1-7 


Slate 




... 




3-2 



APPLIED MECHANICS. 



657 



TABLE XXL— Moduli of Compressibility 



Substance. 


Modulus of 

Compressibility 

in Pounds per 

Square Inch. 


Temperature. 


Authority. 


Steel 


21-6 X 106 




Amagat. 


Steel ... . ... 


26-7 X 106 




Thomson's "Elasticity." 


Iron 


21-1 X 106 




n » 


Brass 


15-3 :< 106 




!' >> 


Copper ... 


17-1 X 106 




Buch;:nan. 


Copper 


24-4 X 106 




Thomson's "Elasticity." 


Delta Metal 


14-4 X 106 




Amagat. 


Lead 


5-3 X 106 




» 


Glass 


5-8 X 106 




" 


Distilled Water... 


3-2 X 105 


15° C. 


Paglianni. 


Alcohol 


1-76 X 105 


0°C. 


Amaury and Deschamps 


Alcohol ... 


1-62 X 10-5 


15° C. 


>> >j 


Ether ... 


1-35 X 10-5 


0° c. 


)» »> 


Ether 


1-15 X 105 


15° C. 


" '» 


Carhon Bisulphide. 


2-32 X 105 


14° C. 


J) ;; 


Glycerine 


5-8'5 X 105 


20-5° C. 


Quincke. 


P<;troleum 


2-11 X 105 


16-5° C. 


Martini. 


Mercury ... 


7-8o X 106 


15° C. 


Amaury and Deschamps. 


Mercury 


4-35 X 106 


0°C. 


Colladon and Sturm. 


Mercury 


3-74 X 106 


0°C. 


Amagat. 



658 



APPLIED MECHANICS. 



TABLE 











Breaking Stress, 








Weiglit 


in lbs. per 


Material. 


Melting 

Point 
(Fahr.). 


Specific 
Gravity.. 


of 

0:16 Cubic 

Foot 

in 


square inch. 












Pounds. 


Tensile. 


Soft Steel (unliardened) \ 


2. 4(X)' 






00,000 to K<0, 000 


Soft Steel (hardened) ( 


to 


7 -So 


490 


120,000 


Cast Steel (imtempeied) t 


3,000^ 






90,0G0 to 153,000 


Cast Steel (tempered) ) 






— 


Steel Plates 








60,000 to 80,000 


Steel Bars 








100,000 to 130,000 


Cast Steel (drawn) 




7 '7 


480 


120,000 


Steel "SVire (English), drawn... 








120,000 to 140,000 


Steel Wire (Common), tern- ) 
pered blue ... ) 




7-4 


460 


330,000 


Steel Pianoforte 'SVire, English 




7( 3 


480 


340,000 


Manganese Steel, Cast 








85,000 


Nickel Steel, unhardened 








75,000 


Xickel Steel, hardened 








190,000 


"Wrought Iron Bars and Bolts^ 








55,000 to 70,000 


Wrought Iron Plates, with 


3.000= 

to 
3 300'' 








fibre 

Wrought Iron Plat-es (across 


7-7 


480 


51,000 


fibre) 








46.000 


Wrought Iron Plates (mean)^ 








48,500 


Castli-on { 


2,0fX)"' 
to2,50ir 


7-2 


450 


14,000 to 30,000 


Aluminium 


1,300= 


2-6 


162 


33.0130 


Aluminium (annealed) 








13. .500 


Brass, YeUow 1 

Brass, Sheet j 


1,700= 


7-8 


490 


17; .500 


to 1,850- 


to 8-4 


to 525 


30,000 


Brass, Tube 








80,000 to 100,000 


Brass, Cast 




8-0 


500 


20,000 


Brass, Wire 








50,000 


Copper, Wrought 


2,0<jO^ 


S-8 


550 


33.000 


Copper, Cast ... 








20,000 


Copper Wue, hard-drawn ... 




8-9 


555 


58.000 


Copper Wire, annealed 








47.000 


Zinc, Ca-st 


750= 


7 


436 


7:5130 


Zinc, Sheet 




72 


4.50 


30.000 


Lead 


615= 


11-4 


712 


1,900 


Lead, Cast 




11-2 


700 


3,000 


Tin ... 


450= 


7-4 


462 


4,600 


Platinum Wire 


3.300'- 


21-5 


1.310 


50.CK30 


Gold ^diawn) 


2,200= 


19-2 


1.200 


38,000 to 41,000 


Silver (drawn) ... 


1,850= 


10-4 


'650 


42.000 


Phosphor Bronze (cast ) 








55;000 


Phosphor Bronze AVire (hard)... 








100,000 to 150,000 


Phosphor Bronze Wire (an- ) 
nealed) ] 








50,000 to 60,000 



AJPPLIEb iMECHANICS. 



659 



XXII. 



Breaking Stress, 

in lbs. per 

square inch. 


Stress -which produces 

Permanent Set, 
in lbs. per square inch. 


Safe Limit of 

Stress, 

in lbs. per square inch. 


Young's 
Modulus 
of Elas- 
ticity, in 


1 

Com- 
pressive. 


Shear. 


Tensile. 


Com- 
pressive. 


Shear. 


Tensile. 


Com- 
pressive. 


Shear. 


millions 

of lbs. per 

square 

inch. 






35,000 




26,500 


17,700 


17,700 


13,000 


30 
30 
30 
36 

29 to 42 

27 
28 
26 
29 


50,000 


50,000 


45,000 
35,000 
56,000 
24,000 


24,000 


20,000 


10,000 


10,000 


7,800 


29 
25 






20,000 


20,000 


15,000 


10,000 


10,000 


7,800 


27 
26 


1 50,000 to 
1 120,000 


28,500 


10,500 


21,000 


8,000 


3,500 


10,500 


2,700 


14 to 23 


10,500 




7,000 




5,000 


3,600 




2,700 


9 


58,000 




4,300 
3,200 


4,000 


3,000 


3,600 


3,200 


2,300 


9-2 

14-2 

15 


7,300 




1,500 












•72 






20,000 




14,500 


10,000 




7,400 


6 
24 
12 
11 
14, 



660 



APPLIED MECHANICS. 



TABLE 











Breaking Stress, 








Weight 


in lbs. per 


Material. 


Melting 

Point 

(Fahr.). 


Specific 
Gravity. 


of 

One Cubic 

Foot 

in 


square inch. 












Pounds. 


Tensile. 


Aluminium Bronze, \ 
Cu95A15/ 




8-25 


515 


60,000 


Aluminium Bronze, \ 
Cu 90 Alio J 




77 


480 


100,000 


Manganese Bronze 








65,000 to 85,000 


Delta Metal (cast) 








47,000 


Delta Metal (rolled) 








74,000 


Muntz Metal 








45,000 


Sterro Metal 








60,000 


Gun Metal 


1,900° 


8-6 


536 


25,000 to 50,000 


Ebony 




1-7 


73 




Oak, European 




•93 


58 


14,500 


Mahogany, Spanish 

Ash 




•85 
•8 


53 
50 


15,000 
17,500 


Pitch Pine 




•7 


44 


12,000 


Red Pine 








10,500 


Birch 




•54 


34 


14,500 


Beech 




•7 


44 


11,500 


Fir, Larch 




•53 


33 


11,000 


Fir, Spruce 




•54 


34 


12,500 


Hornbeam 




•76 


47 


15,000 


Teak, Indian 




•78 


49 


15,000 


Lancewood ... 




•95 


59 


20,000 


Elm, British 




•55 


34 


14,000 


Lignum Vitse j 




•65 
to 1-33 


41 to 83 


16,000 


Sycamore 








13,000 


Cedar of Lebanon 




•59 


37 


11,400 


Granite ... 




2-7 


170 




Marble 




2-8 


175 


700 


Limestone 




2-8 


175 




Sandstone 




2-3 


144 


800 


Slate 




2-8 


175 


11,000 


Brick, Bed I 

Brick, Fire 1 




2 to 2-2 


125 to 137 


300 










Brickwork 




1-8 


112 




Concrete... ... ... -j 




1-9 
to 2-2 


119 to 137 




Leather .. 








4,000 


Hemp Eope (in ordinary state) 




1^3 


— 


10,000 


Glass, Plate 




2^7 


170 


2,700 


Ice 




•917 


57 




Quartz Fibre (Professor Boys') 








140,000 



APPLIED MECHANICS. 



661 



XXII. [continued). 



Breaking Stress, 

in lbs. per 

square inch. 


Stress which produces 

Permanent Set, 
in lbs. per square inch. 


Safe Limit of 

Stress, 

in lbs. per square inch. 


Young's 
Modulus 
of Elas- 
ticity, in 
million.s 
of lbs. per 
square 
inch. 


Com- 
pressive. 


Shear. 


Tensile. 


Com- 
pressive. 


Shear. 


Tensile. 


Com- 
pressive. 


Shear. 




' 


17,000 
50,000 












12 
13 

10 


18,000 
10,000 

6,000 


2,300 
650 














1-5 
1-4 
1-6 
1-4 
1-65 

1-35 

1-1 

16 


12,000 

7,000 
















2-4 


10,000 
















1-04 
•48 


17,000 

6,000 

5,000 

5,000 

15,000 

800 

2,000 

500 
















7 
6 

2 
13 


2,000 








! 
1 
























•025 


26,000 
















8 
8^5 



663 



APPLIED MECHANICS. 













LOGARITHMS. 






TABLE 





1 2 1 3 


4 


5 


6 


7 


S 9 

1 


12 3 4 5 6 


7 8 9 








:i': '.-V. 


:v-c 


:■::. 


::.:.|::-* 


4 SI.:- .: _ 


~'~- 


12 
13 


;: : 


\z\ . '^ X 


'.'-". : : - 


- - - 






4 s 11 15 15 :;:;.. 5 S-} 34 

3 7 10 14 17 21-24 2S 31 
3 6 10 13 16 1&|23 2-5 29 


;r 


rl " 7 




1 . - ■ - ' ' 


r: "^ 


2945 


:- M 17£2 


3 D 9 12 15 IS 
3 6 S 11 14 17 
3 5 S 11 13 l!3 


21 24 27 
20 22 2.5 
IS 21 24 


il 


-;.? 


-:-' X\- --:- 


:,:; -■'- -;: 


iS ! Ecli 


2 5 7 10 12 15 
2 5 7 9 12 14 
2 4 7 9 11 IS 


17 20 22 
16 19 21 
16 IS 20 


'-'■ 




- '^ '■ 




;::■: f-f: ::: 


2 4 6 S 11 12 


15 17 19 


21 
22 


:^-z 


n^- ;':• :'-; 


:-;- :i-: ::^7 


SocOi 

3747 


So 79 S^^S 
3766 3784 


2 4 6 8 10 li 
2 4 6 ; 8 10 12 

2 4 6| 7 9 11 


14 16 IS 
14 15 17 
13 15 17 


25 
2o 


::: 


i^J- iii^ i 


' ! :' 7 ', 


i-J ■-:■ 


2 4 5 7 9 11 
2 3 5 7 9 10 

2 3 5 } T 8 10 


12 14 16 
12 14 15 
11 13 J 5 


27 
29 


4r'. _i 


:j': M: ::- 


i;- :;;• :-_i 


::-: 


_-_- ^7.-7 


2 3 5 

2 3 5 

13 4 


6 8 9 
6 8 9 
6 7 9 


11 13 14 
11 12 14 
10 12 13 


30 4":: 


.-: 4VO 


13 4 


6 7 9 


10 11 13 


31 -.::-. 

32 ::--- 

33 :::: 


-: ; -_ -:,: 


:':: :':2 sias 


.5€4- 




1 3 4 
1 3 4 
13 4 


6 7 8 
5 7 8 
5 6 8 


10 11 12 
9 11 12 
9 10 12 


34 

3-5 
36 


;^- 




13 4 
12 4 

12 4 


5 6 8 

5 6 7 
5 6 7 


9 10 11 

9 10 11 

8 10 11 


37 
35 
39 


■-^ = 


-I- ^Ht Hi: 


w. 


^-;- :'-- 




m M 


12 3 
1 2 3 

12 3 


5 6 7 
5 6 7 

4 5 7 


8 9 10 
3 9 10 
S 9 10 


i'J 


::;-_ : .: 




'.'-'.' •;::- 


: : : 4 5 6 


8 9 10 


■i"'. 


"-^ 


'-\ -:\ 


i " ' - 




-..: ;.- _.-- 


12 3 
12 3 
I 2 3 


4 5 6 
4 5 6 

4 5 6 


7 8 9 
7 8 9 
7 8 9 


46 




r" ;;^ 


-,!^:-; 


: " ~ ^ - ; 


T: -:- -^: 


1 1 £ 
1 2 £ 
12 3 


4 5 6 
4 5 6 
4 5 6 


T 8 9 

7 8 9 


47 

4? 

11 


--'-"'- 


:^rr ^ 


';; \ j:- 


-- -:: 


;^^ ''n ^;-^ 


12 3 
12 3 

1 2 3 


4 5 5 
4 4 5 
4 4 5 


6 7 i 
6 7 S 
6 7 S 


5-: 






1 


--:- 


' .-=. 


-:'; 


--^ -: - 


1 2 3 


3 4 5 


6 7 5 


51 
52 
53 




"_'_ ' 


:-:: It!; 


1 2 3 

12 2 
1 2 2 


3 4 5 
3 4 5 
3 4 5 


6 7 i 
6 7 7 
6 6 7 


"ii^ 


-::- - :.: 


" : T : " : : 


' - 


--; -.: - 


: ■: 2 


3 4 5 


6 6 7 



APPLIED MECHANICS. 



663 



XXITT. 



LOGARITHMS. 



! 

55 

56 

57 

^i 

'59 
60 
61 

62 
63 
64 

65 

66 

67 

69 ! 
70 
71 ' 

72 
73 

75 

76 
77 
78 

79 
80 
81 

82 
83 
84 

85 

86 
87 
88 

89 
90 
91 

92 
93 

94 

~95~ 

96 
97 
98 

89 




7404 

7482 
7559 
7634^ 

7709 
7782 
7853 

7924 
7993 
8062 

8129 

8195 
8261 
8325 

8388 
8451 
8513 

8573 
8633 
S692 

8751 

8808 
88G5 
8921 

8976 
9031 
9085 

9138 
9191 
9243 

9294 

9345 
9395 
9445 

9494 
9542 
9500 

9638 
9685 
9731 

9777 

9823 
9SGS 
9912 

9956 


1 

7412 


2 

7419 


3 


4 

7435 

7513 
7589 
76G4 

7738 
7810 
7882 

7952 
8021 
8089 

8156 

8222 
8287 
8351 


5 


6 

7451 

7528 
7604 
7079 

7752 
7825 
7S96 

7966 
8035 
8102 

8169 


7 


8 


9 


12 3 


4 5 6 


7 8 9 


7427 

7505 
7582 
7657 

7731 
7803 

7875 


7443 


7459 


7466 

7543 
7619 
7694 


7474 


1 2 2 


3 4 5 


5 6 7 


7490 
7566 
7642 

7716 
7789 
7860 

7931 
8000 
8069 


7497 
7574 
7649 


7520 
7597 
7672 


7536 
7612 
7686 


7551 
7G27 
7701 


1 2 2 
12 2 
1 1 2 


3 4 5 
3 4 5 
3 4 4 


5 6 7 
5 6 7 
5 6 7 


7723 
7796 
7868 


7745 
7818 
7889 


7760 
7832 
7903 


7767 
7839 
7910 


7774 
7846 
7917 


1 1 2 
1 1 2 
112 


3 4 4 
3 4 4 
3 4 4 


5 6 7 
5 6 6 
5 6 6 


7938 
8007 
8075 


7945 
8014 

S0S2 

8149 


7959 
8028 
8096 

8162 

822S 
8293 
8357 

8420 
8482 
8543 

8603 
8663 
8722 

8779 

8837 
8893 
8949 

9004 
9058 
9112 

9165 
9217 
9269 

9320 

9370 
9420 
9469 

9518 
9566 
9614 


7973 

8041 
8109 


7980 
8048 
8116 


7987 
8055 
8122 


1 1 2 
112 

1 1 2 


3 3 4 
3 3 4 
3 3 4 


5 6 6 
5 5 6 

5 5 6 


8136 

8-:o2 

8267 
8331 

8395 
8457 
8519 

8579 
8639 
8698 

8756 

8814 
8S71 
8927 

8982 
9036 
9090 

9143 
9196 
9243 

9299 


8142 

8209 
8274 
8338 


8176 


8182 


8189 


1 1 2 


3 3 4 


5 5 6 


8215 
82S0 
8344 

8407 
8470 
8531 


8235 
8299 
8363 


8241 
8306 
8370 


8248 
8312 
8376 


8254 
8319 
8382 


112 
1 1 2 
1 1 2 


3 3 4 
3 3 4 
3 3 4 


5 5 6 
5 5 6 
4 5 6 


8401 
8463 
8525 


8414 
8476 
8537 

8597 
8657 
8716 

8774 

8831 
8887 
8943 

8998 
9053 
9106 

9159 
9212 
9263 

9315 

9365 
9415 
9465 

9513 
9562 
9609 

9657 
9703 

9750 

9795 

9S41 
9S86 
9930 

9974 


8426 
8488 
8549 

8609 
8669 
S7S7 


8432 
8494 
8555 


8439 
8500 
8561 


8445 
8506 
8567 


1 1 2 
1 1 2 
112 


2 3 4 
2 3 4 
2 3 4 


4 5 6 
4 5 6 
4 5 5 


8585 
8645 
8704 


8591 
8651 
8710 


8615 
8675 
8733 


8621 
8681 
8739 


8627 
8686 
8745 


112 
112 
112 


2 3 4 
2 3 4 
2 3 4 


4 5 5 
4 5 5 
4 5 5 


8762 


8768 

8825 
8882 
8938 

8993 
9047 
9101 


87S5 


8791 


8797 


8802 


1 1 2 


2 3 3 


4 5 5 


8820 
8876 
8932 


8842 
8899 
8954 

9009 
9003 
9117 


8848 
8904 
8960 


8854 
8910 
8965 


8859 
8915 
8971 

9025 
9079 
9133 


112 

1 1 2 
112 

1 1 2 
1 1 2 
1 1 2 


2 3 3 
2 3 3 
2 3 3 

2 3 3 
2 3 3 
2 3 3 


4 5 5 
4 4 5 
4 4 5 

4 4 5 
4 4 5 
4 4 5 


8987 
9042 
9096 


9015 
9069 
9122 


9020 
9074 
9128 


9149 
9201 
9253 


9154 
9206 
9258 


9170 
9222 
9274 

9325 

9375 
9425 

9474 

9523 
9571 
9619 

9666 
9713 
9759 


9175 
9227 
9279 


9180 
9232 

9284 


9186 
9238 
9289 


1 1 2 
112 
112 

1 1 2 

112 
Oil 
Oil 


2 3 3 
2 3 3 
2 3 3 

2 3 3 

2 3 3 
2 2 3 
2 2 3 


4 4 5 
4 4 5 
4 4 5 

4 4 5 

4 4 5 
3 4 4 
3 4 4 


9304 


9309 


9330 


9335 


9340 

9390 
9440 
9489 

9538 
9586 
9633 

9680 

9727 
9773 

9818 


9350 
9400 
9450 

0499 
9547 
9-595 

9643 
9GS9 
9736 

S7S2 

9827 
9S72 
9917 

9961 


9355 
9405 
9455 


9360 
9410 
9460 


9380 
9430 
9479 


9385 
9435 
9484 


9504 
9552 
9600 


9509 
9557 
9605 

9G52 
9G99 
9745 

9791 


9528 
9576 
9624 


9533 
9581 
9628 


1 1 
1 1 
Oil 


2 2 3 
2 2 3 
2 2 3 


3 4 4 
3 4 4 
3 4 4 


9647 
9694 
9741 


9661 
9703 
9754 

9800 

9S45 
9S90 
9934 

9978 


9671 
9717 
9763 


9675 
9722 
9768 


Oil 
Oil 
Oil 


2 2 3 
2 2 3 
2 2 3 


3 4 4 
3 4 4 
3 4 4 


9786 

9832 
9877 
9921 


9805 

9S50 
9894 
9939 


9S09 


9814 


1 1 


2 2 3 


3 4 4 


9836 
9881 
9926 

9969 


9854 
9S99 
9943 


9859 
9903 
9948 


9SG3 
9908 
9952 


1 1 

oil 

1 1 
1 1 


2 2 3 
2 2 3 
2 2 3 

2 2 3 


3 4 4 
3 4 4 
3 4 4 

3 3 4 


9965 


9983 


9987 


9991 


9996 



664 



APPLIED MECHA!^C8. 



ANTILOGAEITHMS. 



TABLE 








1 


2 


3 


4 1 5 1 6 


7 S 9 


123 '4 56789 


■oo 

■01 
•02 
•03 




1 "-L 


K'-32 
1076 


i«>;'7 

I'JSO 
1054 


Iv* ; 101-2 ■ lOli 


V:\^. 10::.' io-;i 


(j t') 11 •_ 122 2 


':T 


1033,1030,10^. 
1(K7 1 i.>.5u ! 10*52 
1'>S1 1 10S4 1 l.>-o 


I'j -:0 , 1042 , li>4;l 
l'>?4 , ltx.7 1 l';»5y 
\<hVi , lO^l j 10»V4 


<) >■ 1 : 1 1 2 2 2 
1 1 L 1 ■- 2 2 
1 1 1 1 2 :: 2 


■85 

•06 


11±2 

114^ 




:::- 


r: ■ 


:;- ' :: 1112 


IIU 
1140 

llo7 


1117 1 Illy 
1143 1146 
1169 1 1172 


Oil 
1 1 

oil 


1 1 2,2 2 2 
112 2 2 2 
112 2 2 2 


■0-7 
•C'8 
1» 


-'■ -.- '^ 






:r;; ;-. ;--. 


'I'M 


11^ i ll;>'» 




_ - : : _ . ; _ - 


- .->.•» 


^=^T ,.,.-,- 


,-,•]••- .-, -T -i 


:.:" :::; :..-•: 


'j 1 1 1 : 2 2 2 £ 


10 




:: - i-.: i-.: 


--"- '.-'-- :_" 


--' ----- ■--:: 


; 1 1 I 1 L 2 2 c 


n 

12 

■13 




1- : iiW: ' ii^r 

L :.: 1324 13^27 




-::: -r '-\. 


,1112 2:223 
1 11 2 22 2 3 
1 11 2 2 2 3 3 


14 
■15 
16 

if 
18 
19 


li-io 


:i; a;; a- 


:-: :^-! :"- 


- HrUS 


; 111 2 2 2 3 s 

1112 2 2 3 3 
1 lil 2 2J2 3 3 


147'c' 
1514 
l->4'.^ 


::'i -:-; :;- 


'-' '. '-' -' 1:70 


1 : ; : :7'>7 ; 1510 

l-:c.S , 1542 l.>45 
1574 i 157S ^ 1541 


1 11 2 2 2 3 3 

1 1 I 2 2 2 3 £ 
1 1 1 2 2 i 3 i 


■20 


I'S: 


: : : - 1 : - : r : 


1 ; : '-'-'.-- 1 :;~ 


I"-: ::-4 :::5 


: 1 1 1 2 2 3 3 3 


•a 

■22 
•23 

■23 
•26 


:'-- 


:-■! ---- i-; 


• -- ''^" -'T- 


1"-. '.'V. :-;:4 


112 2 2 
1 12 2 2 

' 1 1|2 2 2 


3 3 3 
3 3 3 
3 3 4 


:--] 


!;B ;iH Wi 


:-- ^: - ^^:: 


'\\. 'V'_ ' S- ). 


•J 1 12 2 2 
1112 2 2 
: 1 12 2 3 


3 3 4 
3 3 4 
3 3 4 


•o- 




':': r r- : T- 


- '. : -- : -- 




; 112 2 3 
112 2 3 

: 112 2 3 


3 3 4 
3 4 4 

3 4 4 


-30 

■31 


Ir:-: 


:;■:■; l/.-l ::':; 


:::^ -'.-'- '.'..'■- 




1 1 ■- - i 4 4 




■^:- -i:^ ::- 


:;;: -:-: t:-: 


-::: -::: -::; 


112 2 3 3 4 4 

1 1 2 2 3,3 4 4 

; 1 1 2 2 3,3 4 4 


-34 

■26 


-^;' 


---^ 


--- -T-r 


^-^ --'--. --\- 


-'\ --H -:; 


1 i 2 2 3 314 4 5 
1 1 2 2 3 3[4 4 5 

112 2 3 3 4 4 5 


-37 
-33 
-39 


r:7^ 


^-•^ 


'-- ':-' --: 


:.;: :;.; :,. 


1 1 2 2 c t 4 4 ; 


■40 


:-:: 


- - 1 : - .' - : - - - 


-; ■ -'^l -:^' 


-.wi ;-:: -: . 


112-1445 5 


■41 
•42 
■43 




--'; -;- r!^^ 


-::; :.:r t.^ 


--:7 >.: >^: 


1 1 2 2 4 4 5 6 


•A4 
■45 
•46 


"':: 


L-'l 


■;' ;:: 


-■77 r i: "':; 


-- ' r;.: :::: 


1 1 2 1 1 4 . 5 t 
1 1 2 £ £ 4 5 5 »j 
1 1 2 3 3 4 1 5 5 e 


•47 
•45 

•49 


-:-7 




':-- '■^- ^:;- 


'y ;:-: 1:-^ 


112 3 3 4 5 5.: 

112 3 4 4 5 6' 

112 3 4 4 - I- ' 



APPLIED MECHANICS. 



665 



XXIV. 



ANTILOGARITHMS. 








1 


2 


3 


4 


5 


6 


7 


8 


9 


12 3 


4 5 6 


7 8 9 


•50 


3162 


3170 


3177 


3184 


3192 


3199 


3206 


3214 


3221 


3228 


1 1 2 


3 4 4 


5 6 7 


•51 
•52 
•53 


3236 
8311 
33SS 


3243 
3319 
3396 


3251 
3327 
3404 


3258 
3334 
3412 


8266 
3342 
3420 


L2.-3 
S350 
3428 


3281 
8857 
3436 


3289 
3865 
3443 


3296 
3878 
3451 


8304 
3381 
3459 


1 2 2 
1 2 2 
1 2 2 


3 4 5 
8 4 5 
3 4 5 


5 6 7 

5 6 7 

6 6 7 


•54 
•55 
•53 


3467 
3548 
3631 


3475 
3556 
3639 


3483 
3565 
3648 


3491 
3573 
3656 


3499 
3581 
8664 


3508 
3589 
3673 


3516 

3597 
3681 


3524 
8606 
3690 


3532 
3614 
3698 


3540 
3622 
3707 


1 2 2 
1 2 2 
1 2 3 


3 4 5 
3 4 5 
8 4 5 


6 6 7 
6 7 7 
6 7 8 


•57 
•58 
•59 

•60 


3715 
3802 
3890 


3724 
3811 
3899 


3788 
3819 
3908 


3741 
3828 
3917 


8750 

3837 
3926 


8758 
3846 
8936 


8767 
8855 
3945 


3776 
3864 
3954 


3784 
8878 
3968 


3793 
8882 
3972 


1 2 3 
1 2 3 
1 2 3 


8 4 5 
4 4 5 
4 5 5 


6 7 8 
6 7 8 
6 7 S 


3981 


3990 


8999 


4009 


4018 


4027 


4086 


4046 


4055 


4064 


1 2 3 


4 5 6 


6 7 8 


•61 
•62 
•63 


4074 
4169 
4266 


4083 
4178 
4276 


4093 
4188 
4285 


4102 
4198 
4295 


4111 

4207 
4305 


4121 
4217 
4315 


4130 

4227 
4325 


4140 
4236 
4335 


4150 
4246 
4845 


4159 
4256 
4355 


12 3 

1 2 3 
1 2 


4 5 6 
4 5 6 
4 5 6 


7 8 9 
7 8 9 
7 8 9 


•64 
•65 
•66 


4365 
4467 
4571 


4375 
4477 
4581 


4385 

4487 
4592 


4395 

4498 
4603 


4406 
4508 
4613 


4416 
4519 
4624 


4426 
4529 
4634 


4436 
4539 
4645 


4446 
4550 
4656 


4457 
4560 
4667 


1 2 3 
1 2 3 
12 3 


4 5 6 
4 5 6 
4 5 6 


7 8 9 
7 8 9 
7 9 10 


•67 
•68 
•69 


4677 
4786 
4898 


4688 
4797 
4909 


4699 
4808 
4920 


4710 
4819 
4932 


4721 
4831 
4943 


4782 

4842 
4955 


4742 
4853 
4966 


4753 
4864 
4977 


4764 
4875 
4989 


4775 
4887 
5000 


12 3 
12 8 
12 3 


4 5 7 

4 6 7 

5 6 7 


8 9 10 
8 9 10 
8 9 10 


•70 


5012 


5023 


5035 


5047 


5058 


5070 


5082 


5093 


5105 


5117 


12 4 


5 6 7 


8 9 11 


•71 
•72 
•73 


5129 

5248 
5370 


5140 
5260 
5383 


5152 
5272 
5395 


5164 
5284 
5408 


5176 
5297 
5420 


5188 
5309 
5433 


5200 
5321 
5445 


5212 
5338 
5458 


5224 
5346 
5470 


5286 
5358 
5483 


12 4 
1 2 4 
1 3 4 


5 6 7 
5 6 7 
5 6 8 


8 10 11 

9 10 11 
9 10 11 


•74 
■75 
•76 


5495 
5623 
5754 


5508 
5636 
5768 


5521 
5649 
5781 


5534 
5662 
5794 


5546 
5675 
5808 


5559 
5689 
5821 


5572 
5702 
5834 


5585 
5715 
5848 


5598 
5728 
5861 


5610 
5741 
5875 


1 3 4 
1 8 4 
1 3 4 


5 6 8 
5 7 8 
5 7 8 


9 10 12 
9 10 12 
9 11 12 


•77 
•78 
•79 


5888 
6026 
6166 


5902 
6039 
6180 


5916 
6053 
6194 


5929 
6067 
6209 


5943 
6031 
6223 


5957 
6095 
6237 


5970 
6109 
6252 


5984 
6124 
6266 


5998 
6138 
6281 


6012 
6152 

6295 


1 3 4 
1 3 4 
13 4 


5 7 8 

6 7 8 
6 7 9 


10 11 12 
10 11 18 
10 11 13 


•80 

•82 
•83 

•84 
•85 
•86 

•87 
•88 
•89 

•92 
•93 


6310 


6324 


6839 


6853 


6368 


6383 


6397 


6412 


6427 


6442 


1 3 4 


6 7 9 


10 12 13 


6457 
6607 
6761 


6471 
6622 
6776 


6486 
6637 
6792 


6501 
6653 
6808 


6516 

6668 
6823 


6531 
6683 
6839 


6546 
6699 
6855 


6561 

6714 
6871 


6577 
6730 
6887 


6592 
6745 
6902 


2 3 5 
2 3 6 
2 3 5 


6 8 9 
6 8 9 
6 8 9 


11 12 14 
11 12 14 
11 13 14 


6918 
7079 

7244 

7413 

7586 
7762 


6934 
7096 
7261 


6950 
7112 

7278 


6966 
7129 
7295 


6982 
7145 
7311 


6998 
7161 
7328 


7015 
7178 
7345 


7031 
7194 
7362 


7047 
7211 
7379 


70d3 
7228 
7896 


2 8 5 
2 3 5 
2 3 5 


6 8 10 

7 8 10 
7 8 10 


11 18 15 

12 18 15 
12 13 15 


7430 
7603 

7780 


7447 
7621 
7798 


7464 
7638 
7816 


7482 
7656 
7834 


7499 

7674 
7852 


7516 
7691 

7870 


7534 

7709 
7889 


7551 

7727 
7907 


7568 
7745 
7925 


2 3 5 
2 4 5 
2 4 5 


7 9 10 
7 9 11 
7 9 11 


12 14 16 

12 14 16 

13 14 16 


7943 


7962 


7980 


7998 


8017 


8035 


8054 


8072 


8091 


8110 


2 4 6 


7 9 11 


18 15 17 


8128 
8318 
8511 


8147 
8337 
8531 


8166 
8356 
8551 


8185 
8875 
8570 


8204 
8395 
8590 


8222 
8414 
8610 


8241 
8433 
8630 


8260 
8453 
8650 


8279 1 8299 
8472 8492 
8670 8690 


2 4 6 
2 4 6 
2 4 6 


8 9 11 
8 10 12 
8 10 12 


18 15 17 
14 15 17 
14 16 18 


•94 
•95 
•96 

•98 
•99 


8710 
8913 
9120 


8730 
8933 
9141 


8750 
8954 
9162 

9376 
9594 
9817 


8770 
8974 
9183 


8790 
8995 
9204 


8810 
9016 
9226 


8831 
9036 
9247 


8851 
9057 
9268 


8872 
9078 
9290 


8892 
9099 
9311 


2 4 6 
2 4 6 
2 4 6 


8 10 12 
8 10 12 
8 11 13 


14 16 18 

15 17 19 
15 17 19 


9333 

9550 
9772 


9354 
9572 
9795 


9397 
9616 
9840 


9419 
9688 
9868 


9441 
9661 
9886 


9402 
9683 
9908 


9484 
9705 
9931 


9506 
9727 
9954 


9528 
9750 
9977 


2 4 7 
2 4 7 
2 5 7 


9 11 18 
9 11 18 
9 11 14 


15 17 20 

16 18 20 
16 18 20 



666 



APPLIED MECHAXICS. 
TAELE XXV. 



Aldr. 


Radiaits. 


Siiifc. 


Ti-r-:.: 


r.:-z^:,TA. 


C<ta3ie. 






0'- 


,-, 


,-, 


(j 


X 


1-0000 


: " - 




l 


-0175 


■0175 


-0175 


57-2900 


-9998 


1-5: 




2 


-0349 


-0349 


-0349 


28-6363 


-9994 


l-5v 




3 


-0524 


-0523 


■0524 


19^11 


-9986 


1-51S4 


v7 


4 


•©€98 


-0698 


•0699 


14-3006 


•9976 


1-5010 


86 


5 


-0S73 


-0872 


-0S75 


114301 


•9962 


14835 


85 


6 


-1047 


-1045 ! 


-1051 


9^5144 


•9945 


1-4661 


84 


7 


-12?f2 


-1219 


-1228 


8-1443 


-9925 


1-4486 


83 


8 


-1396 


-1392 


-1405 


71154 


-9903 


1-4312 


82 


9 


-1571 


-1-564 


-1584 


6-313S 


-9877 


1-4137 


?1 


10 


-1745 


-1736 


-1763 


5-6713 


-9848 


1-Sr:: 


- 


11 


-1920 


-1908 ! 


-1944 


5-1446 


-9816 


1-s:^: 


■ ; 


12 


-2094 


-2079 ; 


-2126 


4-7046 


-9781 


- _ 


Jn 


13 


-2269 


-2250 


•2309 


4-3315 


■9744 


1 ; ^ ; ; 


77 


U 


-2443 


-2419 


-2493 


4-0108 


-9703 


1 ^-^■■i 


76 


15 


-2618 


-2588 


"2679 


37321 


•9659 


1-3090 


75 


16 


-2793 


-2756 


-2867 


3-4874 


-9613 


1-2915 


74 


17 


•2967 


-2924 


-3<»7 


3-2709 


-9563 


1-2741 


73 


18 


-3142 


-3090 


-3249 


3^»^V^ 


-9511 


1-2566 


72 


19 


-3316 


-3256 


-3443 


2-9042 


•9455 


1-2392 


71 


20 


-3491 


-3420 


-3640 


2-7475 


-9397 i 


1-2217 


70 


21 


-3665 


-3584 


-3839 


2-6051 


•9336 


1-2'''^^- 


AQ 


22 


-3840 


-3746 


-4040 


2-4751 


-9272 






23 


-4014 


-3907 


-4245 


2-3559 


-9205 


1 1 . : - 


' 


24 


-4189 


-4067 


-4452 


2-2460 


-9135 


111 




25 


-4363 


-4226 


-4663 


2-1445 


-9063 


ii:. 




26 


-4538 


-4384 


-4877 


2-0503 


-St- 


. - . . 


^ 


27 


-4712 


-4540 


-5095 


1-9626 


-s-i: 






28 


-4887 


-4695 


-5317 


18807 


-.s^: : 


- . '. 




29 


-5061 


-4848 


-5543 


1-8040 


-S-. ; 






30 


5236 


-5000 


-5774 


1-7321 


-*•: : . 


1 \ -.' . 




31 


-5411 


-5150 


-6009 


1-6643 


-8572 


• 


' r 


32 


-5585 


-5299 


-6249 


l-6«03 


-84?<? 




' > 


33 


-5760 


-5446 


-6494 


1-5399 


-Sc^- 


;^s 


57 


34 


-5934 


-5592 


-6745 


1-4826 


-f: 


-74 


56 


35 


-6109 


-5736 


-7002 


1-4-2S1 




55 


36 


-6283 


-5878 


-7265 


1-3764 




- . ■ 


54 


37 


-6458 


-6018 


-7536 


1-3270 


7 ; 


- . ' 


53 


38 


-6632 


-6157 


-7813 


1-2799 


-7»sO 


' 


52 


39 


-6807 


-6293 


-8098 


1-2349 


-7771 


- ; 1 


51 


40 


-6981 


-6428 


-8391 


1-1918 


-7660 


•S727 


.30 


41 


-7156 


-6561 


-8693 


1-1504 


-7547 


-8552 


49 


42 


-7330 


-6691 


-9004 


1-1106 


-7431 


-8378 


48 


43 


-7505 


-6820 


-9325 


1-0724 


-7314 


-8-203 


47 


44 


-7679 


-6947 


-9657 


1^>35 


-7193 


-8<»-29 


46 


45 


-7854 


-7071 


1-OGOO 


1-0000 


-7071 


-7854 


45 






Coaue 


(Vtss^eait 


TsngaA 


Sfee 


Badia-D? 


Ai^-k 



INDEX 



The References are to Paj's. 



Absorption Dynamoineters, 233 
Acceleration, 13, 262 

• Angular, 25 

Linear, 13 

in Mechanism, 583 

inS.H.M., 548 

Accumulator, 187 

Advantage, Mechanical, 59, 88 

Air-vessel, 506 

Algebra, 1 

Alloys of Copper, 288 

of Iron and Manganese, 287 

of Iron and Nickel, 287 

Aluminium, 285, 289 

Bronze, 289 

Amperes, 92 
Amplitude, 20 

■ of Vibration, 565 

Analogies, Linear and Angular Motion, 593 
Angle, Measurement in Degrees, Radians, 
23 

of Repose, 110, 112, 125 

Sine, Cosine, Tangent of, 23 

-of Twist, 225, 349 

Angular Acceleration, 25 

• Motion, 586 

Ratio, 571 

Velocity, 25 

Aqueduct, 208 
Arch, 113, 891, 457 

Least Thrust at Keystone of, 165 

Line of Resistance of, 163 

Load Carried by, 163 

■ Masonry, 166 

■ Metal, 478 

■ Middle Third of Section, 163 

Rib of Iron, 166 

Thrust at Crown of, 163 

Arched Rib, 162 
Area, Centre of, 136 

of Curve by Integration, 17 

of Curve by Planimeter, 7 

of Irregular Figure, 10 

— ;— Moment of Inertia of, 136 

— — Projection of, 27 

Areas of Figures, 6 

Armstrong's Hydraulic Crane, 194 

Ash, 278 

Attwood's Machine, 64, 245 

Average Curvature, 22 

Space, Time, 272, 274 

- — Velocity, 12 



The References are to Pagfs. 

Axial Flow Turbine, 531 
Axis, Neutral, 381 

of Suspension, 560 

Axles, Railway, 83 

Ayrton- Perry Coupling, 239 

Ayrton and Perry's Dynamometer, 225 



Babbit's Metal, 289 

Bailey's Testing Machine, 294 

Balance, Chemical, 118 

Compensation of, 559 

of Watch, 558 

Spring, 40 

Balancing of Hoists, 201, 203, 204, 205 

of Locomotive, 609 

of Machines, 606 

of Reciprocating Body, 611 

Bail-Bearings, 69, 86 
Ballistic Pendulum, 499 
Barge, 208 
Barker's Mill, 486 
Basin, Water in, 539 
Bauschinger's Experiments, 310 
Beam, 127 

Bending Moment Diagrams, Six 

Cases, 415, 428 

Cast Iron, Weight of, 9 

Curvature of a, 386 

Fixed at the Ends, 413, 443 

Not Rectangular in Section, 417 

Shear Stress in, 459 

Shearing Force in, 414, 427 

Stiffness of, 431 

of Uniform Depth, 446 

of Uniform Strength, 419 

Well-known Rules about, 410 

Without Compression, 471 

Bear, Punching, 179 
Bearing, Diameter of, 68 

Friction at, 68 

Bearings, Ball, 69, 86 

of Fans, Centrifugal Pumps, Dy. 

namos, 70 
Beech, 278 

Bell-mouthed Pipe, 522 
Belt, Centrifugal Force on, 327 

Length of Crossed, 26 

Length of Open, 26 

Slip of, 86 

Belting, Creep in, 86 
Strength of, 231 



668 



APPLIED MECHANICS. 



The Eeferenees are to Pages. 

Belts, Difference of Pulls in, 226 
Bending, 380 

and Crushing, 457 

Moment, 128, 383 

Moment in Beams (more difficult 

portion), 441 
Moment Diagrams, Six Cases, 414, 

427 

of Prism 369 

of Struts, 464 

Theory of, 318 j 

Unsymmetrical, 387 

Bends in Pipes, 5193 
Bevel Gearing, 36, 95, 102 
Bicycle Problems, 49 

Resistance to Motion of, 50 

Bicyclist, 33 
Bitilar Suspension, 564 
Blue Prints, 2 
Blocic, Pulley, 98 

Snatch, 102 

Blocks and Tackle, 98 

Board of Trade Rules for Railway Girders, 

305, 402 
Body Turning about an Axis, 114 
Boiler, Bursting of, 181 

Cylindrical, 320 

Spherical, 320 

Strength of, 319 

- — Testing of Water-tube, 93 
Bolts, 100 

Tightening of, 311 

Boundary Condition, 372 

Boussinesq, M., 379 

Boy's, Prof., Quartz Fibres, 299 

Braced Triangular Pier, 148 

Bracket, 459 

Brake Block, 240 

Horse-power, 91, 95 

Knot, 236 

Brass, 288 

Weight of, 9 

Bricks, 276 
Bridge, Forth, 299 

Suspension, 161 

Bronze, 288 

Brotherhood's Water-pressure Engine, 191 

Bulk, Diminution in, 316 

Modulus of Elasticity of, 316 

Bull Engine, important example, 257, 

269 
Bullet, 14, 79, 500 
Bursting of Boiler, 181 
Buttress Thread, 101 
Buttresses, 166, 458 



Calculus, Differential and Integral, 15 
Calorific Power of Coal-Gas, 91 
Cam, 37 
Canal, 170, 206 
Candle. 496 



The References are to Pages. 

Cannon, Momentum of, 486 
Carbon, 280, 281, 285 
Case-Hardening of Wrought Iron, 285 
Castings, Chilled, 183, 284 

Cooling of, 282 

Internal Strains in, 282 

Malleable, 284 

Softening of Hard, 281 

Strain in, 183 

Cast Iron, 280 

Density of, 281 

Shell, 173 

Toughening of, 284 

Weight of, 9 

Cast Steel, 183 
Catenary, 169 
Cauchy's Theorem, 363 
Cedar, 278 
Cement, 276 
Centre of Area, 136 

of Fluid Pressure, 215 

of Gravity, 6, 136 

of Inertia, 136 

Instantaneous, 575 

of Ma.ss, 135 

of Oscillation, 560 

of Percussion, 499, 560 

of Pyramid or Cone, 9 

Centrifugal Force, 597, 606 

Force in Belts or Ropes, 327, 005 

Pump, 33, 264, 540 

Centripetal Force, 603 

Acceleration, 603 

C.G.S. System of Units, 267 
Chain Gearing, 232, 572 

Hanging, 167 

Horizontal Pull of, 1C.2 

Loaded, 161 

Chains, Admiralty Rule for, 315 
Cliange Wheels, 101 
Charcoal Iron, 284 
Chemical Balance, 118 

• Energy, 42 

Chemistry, 290 
Chilled Castings, 183 
Chipping, 488 ' 

and Filing, 56 

- — Hammer, 264 

Chisel, Tempering of, 2?1 

Chree, Dr., 371, 380 

Chromium, 287 

Chronometer, 649 

Chuck, Elliptic, 577 

Circle, Circumference of, 

Circuital, 30 

Circular Stream Lines, 541 

Clinure, 29, 122 

Clock, 38 

Closed Polygon, 120 

Coal, Energy of 1 lb. of, 42, 43 

Co-efficient of Friction, 65 

of Friction, Alteration of, 75, 236 

of Viscosity, 76 

Collars, Leather, 170, 176 
Colours in Tempering, 291, 292 



INDEX. 



669 



The References are to Pages. 

Columns, 473 

Combination of S.H.M.'s, 555 

Combined Bending and Twisting, 353 

Commercial Tests, 306 

Compound Interest Law, 20, 99, 230 

Pendulum, 559 

Wheel and Axle, 102 

Concrete, 183, 277 

Conduit, 516 

Cone, Area of Surface of, 8 

Rolling of a, 577 

Surface of, 8 

Coned Pulley, 37 

Cones, Stepped, 26 

Conical Pendulum, 54;6 

Confluent, 30 

Connecting Rod and Crank, 222 

Construction of Gun, 330 

Materials Used in, 275 

Contracted Vein, 533 
Contraction of Section , 306 
Cooling of Castings, 282 
Copper, 287 

Wire, 289 

Cores, 10, 282 
Cork, Floating, 547 
Corrugated Flue, 326 
Cosine of an Angle, 23 
Cottars, 101 
Cotton Press, 184 
Couple, 127 
Coupling, 225 
— Ayrton-Perry, 239 

Dynamometer, 96 

Flange, 227 

Rod, 471 

Crab, 102 
Crane, 88 

Double-Powered, 195 

Hook, 468 

Hydraulic, 193 

Crank and Connecting-Rod, 222 

Overhung, 355 

Creep in Belting, 86 
Creeping, 299 
Cricket Bat, 500 
Critical Velocity, 83 
Crosshead of Engine, 113, 547 
Crushing and Bending, 457 
Cupola, 282 
Curvature of a Beam, 386, 405 

Definition of, 22 

Radius of, 22 

of Steel Spring, 388 

Curve of Cosines, 555 

Elastic, 390 

Integral of, 18 

JLiOgarithmic, 568 

of Sines, 78, 555, 565, 568 

Curved Rollers, 522 
Curves, Damping, 78 

Trochoidal, 573 

Cutting of Metals, 45 
Cylinder, Rotating, 369 
Thick, 183, 323 



The Referenceslare to Pages. 



Damping Curves, 78 

of Vibration, 566 

D'Arcy's Experiments, 85 

Dash-pots, 239 

Deadloads, 305 

Definition of a Radian, 23 

Definitions of Trigonometrical Ratios, 23 

Deflection due to Shearing, 334, 460 

■ in Beams, 407, 429, 431, 460 

Delta Metal, 288 
Descriptive Geometry, 121 
Diagonal Pieces, 162 

of Velocities, 583 

Diagram of Accelerations, 583 

of Bending Moment, six cases, 413- 

427 
Diagrams of Shearing Force, 427 
Differential Co-efficient, 16 

Pulley Block, Efficiency of, 104 

Pulley Block, 103 

Diminution of Swing, 563 '■ 
Displacement of a Ship, 55 

Relative and Absolute, 35 

Distance of Rubbing, 68 

Draining of Fields, 516 

Drawing Office, 1 

Drill, Watchmaker's, 291 

Dunkerley, Mr., Stability of Shafts, 477 

dy 

dx,' 

d-^ll 



15 



dx^ 



IS 



Dynamo Machine, 43, 92 
Dynamometer Coupling, 96 

Fronde's, Thorneycroft, 234 

Hefner- Alteneck, 235, 240 

■ Rafiard's, 86 

■ Transmission, 92 

Transmission and Absorption, 233 



Earth, Pressure of, 167 

Rankiue's Rule for, 349 

Earthquake, 497-617 
Eccentric, 222 
Economical Efficiency, 89 
Economy, General, 339 ^ 
Eddies, 518 
Effect of Friction, 60] 
Efficiency, 95 

Economical, 89 

■ Hydraulic, 529 

of Screw Jack, 110 

of Turbine, 522 

Elastic Curve, 390 

Elasticity, More Difficult Theory of, 361 

of Bulk, Modulus of, 317 

Shearing, 333 

Young's Modulus of, 299 



670 



APPLIED MECHAVICS. 



The References art to Pages, 

Electric Energy, 92 

Lamps, 43 

Motor, 42 

Supply Station, 92, 94 

Electrical Horse-i>ower, 92, 95 
Electricity, 290 
Ellipse, Area of; S 
Elliptic Trammel, 577 

Chuck, 577 

Klui, 27S 

Energy, Chemical, 42 

Indestructibilitv of. 59 

Kinetic, Potential, 4<3, 242, 505, 5U 

Loss of, due to Friction, 6T 

Loss of, in Chanse of Strain. 303 

Lost, Fluid in Pipe, 53, S4 

of 1 lb. of Coal, 42-43 

Pressure, 50-5-511 

in a Rotating Body, 247, 254 

Storage of, in Fluids, 326 

Store of, ISS 

Stored up in any ilaehine, 253 

Strain, S5 

of Straineii Spring, 41 

Transmission of, Hydraulically, 170 

Waste of, in Conductors, 43 

of Waterfall, 59 

Engine, Bull, 257 

Hero's, 4S6 

Steam, Gas, Oil, 43 

Water Pressure, 1S9-191 

Willans', 611 

English, Colonel, 55 
Epicyclic Train, 35 
Epicycloids, 573 
Equilibrant, 30 

of Forces in One Plane, 122 

Equilibrium, Conditions of, 121 

in One Position, 104 

Equi-poUent Loads, Wo 
Equi-potential Surfaces, 218 
Errors of Observation, 62 
Escapements, 554 
E wing's. Prof., Extensometer, 294 
Expansion, Cubical, Co-efficients of, 5 

Linear, Co-efficients of. 4 

of Girder, 133 

Experiments by Bauschinger, 310 

bv D Arcy, S5 

bV Fairbairn, 309 

bV JI. Tres;-a, 3*31 

bv Revncl'is, Prof. O. , SI 

bV Tower, iL B., 73. SI 

by Wohler, 305, 309 

on Attwoci's Machine, 245 

on Balancing, 60S 

on Crane, &5 

on Deflection of Beams, 431 

on Fluid Friction, 77, 79 

on Forces at a Pin Joint, 151, 152 

on Friction, 00 

on Friction of Cords, Belts, 22S 

on Friction of a Machine, &3 

on Friction and Speed, 71, 72 

en Gu-s. 14 



The RtftTtrMus are to Pa^^s. 

Experiments on Hydraulic Jack, 175 

on Inclined Plane, 104 

on Jack, 63 

on L"aded Links, 160 

on Moments, 115 

on Shearing, 332 

on Strain, 293 

on Triangle of Forces, 56 

on Twisting of Wires. 349 

to Find Co'^ffideni of Friction, 67 

to Find Energy in Rotating B>ly, 

to Find M-.-«dulu3 of Rigidity, 5o3 

\vith Discs in Fluids, 7S 

— ■ "wltli Gas-engine, 91 

"witli Oil-engine. 9i 

wirh Puller Block, 103 

with Small Carriage, 262 

Extens'^meter, Prof. Ewiog's, 294 



Factor, Load, 94 

of Safety. 304 

Fairl-aims Experiments, 309 

FaUing Weight, Work Done by, 39 

Fatigue, 559 

Ferguson's Paradox, 36 

Fir, White, 277 

Firwoods, 2i7 

Flat Plate, 330 

Flexible Pipe, 197 

Flexmal Rigidity, 634 

Flexure, Non-uniform, 378 

Flow of Gas from Orifice, 537 

of Metals, 335 

of Water, Sudden Stoppage of; 192 

of Water through Notch, 514, 536 

Flue, Corrugated, 331 
Fluid Friction, 76 

Friction, Cause of, 80 

Fluids in Motion, 505 

Steady Motion in, 532 

Storage of Energv in, 326 

Whirling, 216 

Force Acting on a B.^y. important 
example, 265 

Centrifugal, 597, 606 

Constant. 267 

Due to Pressure of Fluids, 214 

Lines of, 217 

Moment of, 115 

of a Blow. 490 

— Of a Blow, lime Average of, 273 

of Friction. 64 

Polvson, 124 

Shearing, 3S3, 3S5 

Space Average of a, 45 

Time Average of a, 45 

Forces Acting at a Point 123 

Graphical Representation of, 30 

in One Plane, Analytical Rule, 121 

in Space, Examples on, 143, 144. 145 

in Stractures, Determination of, 156 



INDEX. 



671 



The References are to Pages. 

Forces Not All in One Plane, 124 

Parallelogram of, 30 

Resultant of, 29 

Triangle of, 30, 56 

Forging, Hydraulic, 170 

Forth Bridge, 299 

Four-Bar Kinematic Chain, 577 

Fracture Surface, 306 

Frame of a Machine, 35 

Framed Structures, 143 

Francis, Mr., 586 .^y- 

Frequency, 20 

Friction and Speed, 71 

at Bearing, 68 

at Different Speeds, 73 

at Joints, 112 

at Leather Collar, 171, 176 

Between Cord and Pulley, 228 

Circles, 113 

Co-efficient of, 65 

Cumulative Eflect of, 99 

Eflfect of, 60 

Experiments on, 60 

- — - Fluid, 76 

Fluid, Experiments of, 77, 78 

Force of, 64 

Internal, 495 

• Laws of Fluid and Solid, 80 

Never Negligible, 248 

of a Pivot, 68, 71 

of Pulley, 58, 248 

• of Railwav Brake, 75 

of Ships, 56 

on Inclined Plane, 107, 108 

Quasi-Solid, 177 

Skin, 54 

Statical, 75 

Wlieels, 69 

Frictional Loss, Reduction of, 68 

Frictionless Hinges, 151 

Froude, 54 

Fronde's Dynamometer, 234, 238 

Fuel Consumption, 93 

Fuller, Prof., 164 

Functions, Graphing of, 21 



Galileo, 244 
Gas-Engine, 43 

Experiment with a, 91 

Gas Flowing from Orifice, 537 
Gauche Polygon, 121, 143 
Gani:,c Pressure, 505 
Gearing Chain, 232, 572 

Spur. Bevel, 95, 102 

General Plane Motion, 575 
Geology, 275 
Geometry, Practical, 1 
German Silver, 289 
Girder, 393 

Bending Moment in. 393, 396 

Board of Trade Rules for, 305 

Calculation of Forces in, 151 



I The References are to Pages. 

I Girder, Diagonal Braces of, 393, 396 

Flanges of, 393, 396 

Railwav, 401 

Rolled," 419, 421 

Shearing Force in, 393, 396 

Web of,' 393, 397 

Glass, 279 

Sudden Cooling of, 279 

Toughened, 280 

Graphical Statics, 1, 149, 151 

Graphing of Functions, 21 

Greenhill, Prof., on Stability of Shafts, 47c 

Grid, Hydrauhc, 212 

Guide Blades, 529 

Gun Experiments, 14 

Lifting of, 199 

Making of, 325 

Metal, 288 

Tensions in, 183 

Wire, 325 

Gyration, Radius of, 138, 560 
Gyrostat, 498 



H 

Hammer, Chipping, 264 

■ Tilt, 500 

Hammering of Iron, 285 
Hanging Chain, 167 
Hardening, Case, 285 

of Steel, 290 

Harmonic Motion, 20 

Law, 20 

Heat, 290 

Energy, Unit of, 42 

Heavy Disc, Vibration of, 568 
Hemp-Packing, 177 
Hero's Engine, 486 
Hinge, Quasi, 166 
Hinged Arch, 114 
Hoists, Balanced, 178 

Balancing of, 201, 203, 204, 205 

Hotel, Mill, Warehouse, 200 

Horse-power, 39, 89 

Brake, 91 

Electrical, 92 

Indicated, 92, 95 

Hydraulic Crane, 193 

Efficiency of Turbine, 529 

Forging, 170 

Grid, 212 

— Intensifier, 186 

Jack, 63, 178 

■ Jack, Efficiency of, 175 

Limestones, 276 

Mean Depth, 85 

Power Company, 94, 187 

Power, Loss of, 53 

Press, 170 

Ram, 501 

Transmission, 52 

Hydraulics, 170 
Hypoeycloids, 573 



672 



APPLIED MECHAXICS. 



The R^faremKia are to Faga. 



I. for Tarioos Sectioiis, 351 
Idea of Law of Depoidenoe of two Vari- 
able Things. 63 

of a Bate, U 

of Slope, 15 

of Veloaty, 12 

Iinpaet, 487, m^ 
Impulse, 9&4 
IncUned ^ane, 99. 101 
-^ — Fricy<m on, 107 
Indiarnbber, 305 
- — Crap, iVf 

Shafts 318 

- — Springy 41 

Indicated Hoise-power, ^5 

Indicator Diagram, 40, 274 

Centre of, 149 

Inertia; 40 

Moment of, 135 

Principal Moments oT, 137 

Role for, 7 

Instantaneous Centre, 57-5 
Inti^;ral of a Carre, IS 
Int^iale, Table of, 16 
Intensi&er, 1S6, 1S8 
Involute of CSide, 573 
Inward Radial Flow Tarblme. 551 
Irc'ii. Alloys of, 287 

Cast, 380 

Cbarooal, 2S4 

Galvanised, 2^7 

Rolling and Hammering, 2S5 

Wroii^t,2&4 

IiT^;uIar Flgore, Approsiniatc Aiea of, 
10 



J 



Jack, Hydraulic or Screw, 63, 178 
Jet Pomp, 516 
Joint, Mas(Hiij, 458 
Joints, Riveted, 336 

wiHi Friction, 112 

Joule, 43 

Joornal, Friction at, 6S 

Journals and Footsteps, 73 



Kalers Pendulum, 560 
Kelvin, Loid, 302 
Kelvin's Analogy, 35S 

lide-Predietlng Machine, 555 

Kerosene, Eneigy o^ 42 

Keys, 101 

Keystone of Arch, 165 

Kinematics, 1 

Kinetic Eneigy, 343, SOaSll 

of Any System. 588 

Kinetics (tf Mechanism, 230 



The Eeferemoesan to Paget. 



Kirchhoff; 366 
Knife Edges, 118 
Knot Br^ 336 
Kohlrausch, 303 



Labonut'ury, Mechanical, 1 

Ladder, Esample on, 134 

Lake of Water, 513 

Landing Stage, 306 

Lareh, 378 

Lathe. ScDew-Cattiug, 101 

Law Connecting Variable Tbin^, 63 

of Crane, 89 

of Friction for a Machine, 63 

cif Moments, 115 

of Worth, 5S 

Laws of Fluid and Solid Friction, SO 
Leather CoDais, 170, 176 

Friction at, 171 

Level Sur&ce, 22 
Lever. 116 
Limestones, 275 

Xatuial HydiauUc, 270 

Li Lie of Resistance, 113, 126, 161, 163 

Projection of, 2r 

Vertica], 22 

Ldnes of Force, 217 
link Motions, 5S4 
Link Polygon, 129 
Links, Loiadtd, 160 
Live Load, 305 
Load Factor, &4 

Rolling, ^.tl 

Loaded Chain, 161 

Links, 160 

Loads, Eqoi-polleut, 366 

Suddenly AppUed, 306 

Loam, ^2 

Local Strengthening, 306 

Lock Nuts, 311 

Locomottve, Poll of, 76 

Logarithmic Curve, 5JS 

Logarithms, 1 

Lf«s of Eneigy, by Impact, 4SS 

of Bneiigy due to Friction, 67 

of Eneigy due to Fluid in Rpe, 84 

of Enei^y in Change of Strain, 303 

of Enei^ per lb. of Water, 51S 

<^ Power at Diffoent Parts of En^nc, 

99 
Love's Treatise on Qasticity, 361 
L):weU Fonunla for Rectangular Notch, 

515, 536 
Lubricant, 66 
Bodyo^ 75 



M, for Various Se<!ti<»is, 251 

of a Wheel, 347, 34S, 254 

Madiine, Bailey's Testipg, 294 



INDEX. 



673 



Tht Rejerences are to Pages. 

Machine, Cotton Baling, 186 

Design, 70 

Kinetic Energy Stored up in, 253 

Riveting, 170 

Machinery, Quick Speed, 530 

Water Pressure, 83 

Machines, Balancing of, 606 
— — Forging, Welding, Punching, Stamp- 
ing, Shearing, 196, 198 

Shearing, 335 

Steadiness of, 252 

Magnet, 564 

Mahogany, 278 

Mainspring of Timekeeper, 554 

Malleable Castings, 284 

Manganese, 284 

Bronze, 289 

Marbles, 275 
Masonry Joint, 459 
Mass, 40 

Centre of, 135 

Materials, Behaviour of, 290 

Used in Construction, 275 

Weights of, 9 

Mathematical Tables, 24 
Mean Depth, Hydraulic, 85 
Mechanical Advantage, 59, 8-8 

Hypothetical, 102 

of Hydraulic Jack, 174 

Mechanism, 220, 570 

Four-Link, 35 

Kinetics of, 220 

Quick Return, 578 

Mechanisms, Acceleration of, 583 

Velocity of, 583 

Memel, 277 
Mensuration, 1, 6 
Metal, 288 

Arches, 478 

Babbit's, 289 

Delta, 289 

Flow of, 298, 301, 335 

Gun, 289 

Muntz, 289 

Sterro, 289 

White, 289 

Mills, Driving of, 515 

Milne, Prof., 401 

Mitis Castings, 287 

Modulus of Elasticity cf Bulk, 317 

of Rigidity, 333, 563 

of Section, 398 

-= — Young's, 299, 305 
Moment, Bending, 128, 384 

Law of, 115 

of a Force, 115 

of Inertia, 135, 136 

of Inertia of a Circle, 141 

of Inertia of Area, 149 

of Inertia of Rectangle, 141 

of Inertia of Sections, 398 

of IMomentuni, 500, 528, 587 

Work done by, 115 

M.,mental Ellipse, 142 
Momentum, 263 



The References are to Poqzs. 
Momentum, Important Example, 274 

of Cannon, 486 

of Shot, 486 

Tangential, 524 

Mortar, 276 

Motion, Fluids in, 505 

of Rotation, 127, 592 

of Translation, 121, 592 

Periodic, 546 

Produced by Blow, 498 

Motions, Link Value, 584 
Moulder, 9 
Moulding, 282 
Muntz Metal, 88 



N 



Neutral Axis, 381 

Surface, 381 

Newton's Second Law, 602, 604 

Nickel, 287, 289 

Non-Redundant Frame, Criterion for, 14S 

Norway Spruce, 277 

Notch, 514 

■ Gauge, 515, 536 

Rectangular, 536 

Triangular, 514, 536 

Nozzles, 513 
Nuts, 311 



Oak, English, 278 

Oil Engines, Experiments with, 9) 

Engines, 43 

Sperm, 66 

Tester, Thurston, 81 

Orifice, 512 

Gas Flowing from, 537 

Rectangular, 535 

Sharp-Edged, 513, 534 

Triangular, 535 

Water from, 512 

Oscillating Cylinder Engine, 578 
Outward Radial flow Turbine, 531 
Overhauling, 96, 104, 109 
Overhung Crank, 355 



Packing Hay, 185 
Parabola, 162, 168 
Parabolic Rib, 162 
Paraboloids of Revolution, 218 
Paradox, Ferguson's, 36 
Parallel Motion, Watt's, C9 
Parallelogram of Forces, 30 
Patterns," 10, 282 
Peaucellier Cell, 578 
Pendulum, Ballistic, 499 500 

Compound, 559 

Conical, 546 



674 



APPLIED MECHANICS. 



The References are to Pages. 

Pendulum, Equivalent Simple, 559 

Impulse given to, 551 

Simple, 242 

Simple, Time of Swing, 551 

Percussion, Centre of, 499 
Periodic Motion, 546, 592 

NotS.H.M., 554 

Time, 20, 546 

— — Time of Balance, 558 
Permanent Axes, 607 

Set in Wire, 300 

Perry's, Mr. James, Syphon, 517 

Phosphor Bronze, 2S8 

Phosjihorus, 284, 287 

PhTsies, 290 

Pianoforte Wire, 289 

Piezometer, 173 

Pig-iron, Pudflling and Refining of, 284 

Pife Driver, 273, 480 

Pin, Resultant Force at, 112 

Pine, Red, 277 

Pins, 101 

Pipe, Bell-Mouthed, 522 

Flexible, 197 

Resistance to Motion of Fluid in a, 

76 

Strength of, 182, 319 

Suddenlv Enlarged, 5 

Wooden, 1S3 

Pipes, Bends in, 519 

Flow of Water in, 53 

Piston-Rod, 309 

Pitch of Screw, 100 

Plane Motion, 576 

Planimeter, 2, 7 

Plastic Elongations, 301 

Plasticity, 301 

Plate, Fiat, 330 

Platform, Weight of, 402 

Plotting on Squared Paper, 62, 80, 89, 91, 

92, 229 
Poisson's Ratio, 343 
Polj-gon, Closed, 120 

Gauche, 121 

Link, 129 

of Forces, 124 

Tnclosed. 124 

Portland Cement. 277 

Potential Eneigy, 40, 242, 505, 511 

Poundage of Steam, 95 

Power, Loss of, at Dififerent Parts of 

Engine, 90 

Misuse of this Expression, 89 

. of a Stream, 515 

Transinissinn V.y Shafts, 224 

Precession of Top, 596 
Press, Cotton, 184 

for Packing Hay, 185 

for Warehouses, 185 

Hand, 185 

Hydraulic, 170 

Pressure Energy, 505, 511 

Gauge, 505 

of a Fluid, 215 

of Earth, 1<37, 348 



The Befereiuxs are to Page*. 

Pressure of Water, 167 

on Immersed Surface, 215 

Principal iloment of Inertia, 139 

Stress, 354 

Stresses, 364 

Principle of Work, 120 

Prints, 10, 282 

Prism, Centre of Gravity of, 9 

of Elliptic Section," 370 

Twisting or Bending of, 309 

Volume of, 9 

Prismatic Bodv, Volume of, 9 
ProjectOe, 245* 
Projection of Art-a, 27 

of Line, 27 

Propeller Shaft, 82, 100 
Propulsion of Sliips, 52, 525 
Protractor, 23 
Puddlins of Pig-Iron, 284 
Pull, How it is Exerted, 292 
Pullev Block, 98 

Block, Efficiency of, OS 

Ccned, 37 

Friction of, 58 

Rim of, 2S3 

Pump, 505 

Centrifugal, 33, 264, .DOS 

Double-Acting Force, 506 

Efficiency of, 43 

Force, 506 

Jet, 516 

Lifting, 505 

Mechanism, 578 

Punchina, 335 
Bear, 179 



Q 

Quantity, Scalar, 29 

Vector, 29 

Quartz Fibres, 299 
Quasi-Hinge, 166 

Rigidity, 497 

S'llid Friction, 177 

Quicklime, 276 

Quick Return Mechanism, 



R 



Rack, 573 
Radial Gears, 584 
Radian, Definition of, 23 
Radius of Curvature, 22, 388 

of Gyration, 136, 2:51, 560 

Ratford's Dynamometer, 80 
Railway Axles, 83 

Brake, Friction of, 75 

Girder, 133, 402 

Ram, Hydraulic, 503 

Rankine', 319 

Rankiue's Rule for Earth, 349 

liate. Algebraical Representation of, 16 

Ratio, Velocity, 59 



INDEX 



675 



The Rejerences are to Prujes. 

Rectangular Notch, 536 

Orifice, 534 

Red Pine, 277 
Redundant Bars, 395 
Refining of Pig-iron, 284 
Regulation of Turbine, 529 
Relative Velocity, 82, 84, 35 

Viscosities, 569 

Resilience, 41, 308 

Compressive, 318 

Shear, 316 

Tensile, 316 

Resistance, Line of, 113, 114, 161, 480 

of a Moving Train, 44 

to Motion of Fluid in. a Pipe, 76 

to Rolling, 85 

Wave, 54 

Resolved Part of a Vector, 29 
Resultant Force, 29, 129 

on Forces in One Plane, 122 

Reynolds, Prof. 0., 73, 78, 81-4, 541 
Rib Arched, 162 

Parabolic, 162 

Rigidity, Flexural, 634 

Modulus of, 883, 563 

Quasi, 497 

Torsional, 634 

Rim of Wheel, Volume of, 9 
River Weaver, 206 
Riveted Joints, 336 

Work, 297 

Riveting Machine, 170 
Rocks, Slaty, Stratified, 275 
Rolled Girder Section, 419, 421 
Rollers, Curved, 572 

for Girder, 133 

Rolling, 106 

and Hammering of Iron, 285 

of a Cone, 577 

Load, 403 

Resistance to, 85 

True, 571 

Roof, Calculation of Forces in, 151, 153 

- — Principal, 133 

Rope, Centrifugal Force in, 322 

Weight of, 315 

Ropes, Wire, 232 

Rotating Body, Energy in, 247 

Cylinder, 368 

Rotation, jMotion of, 127 
Rubbing, Distance of, 68 
Rules for Beams, 410 

for Deflection of Beams, 431 

for Flow of Water through Notch, 

515, 536 

for Periodic Time in S.H.M., 549 

for Simple Pendulum, 551 

for Struts, 466 

for Strength of Pipe, 182 

Rupert's Drop, 183, 280 



Safety, Factor of, 30 



The References are to Pages. 

Safety-valve, 116 
Sandstones, 275 

Green, Dry, 282 : 

Scalar Quantity, 29 
Scraped Surfaces, 73 
Screw, The, 99 
Cutting Lathe, 101 

Jack, 63 

Jack, Efficiency of, 110 

Piles, 101 

Pitch of, 100 

Propeller, 101 

Propeller Blades, 289 

Square-Threaded, 109 

Thread, Whitvvorth, Selleis, Buttress-. 

101 
Seasoning of Timber, 101 
Section, Change of Shape of, 374 
Sections of Structures, 159 
Seizing, 73 
Sellers' Thread, 101 
Sense of Vector, 29 
Shaft, Hollow Round, 354 

Indiarubber, 318 

Overtwisted, 290 

Section, au Equilateral Triangle, 37T 

Shafts of Various Sections, 357 

Stability of, 475 

Strength of, 851, 852 

Subjected to Twisting and Bending. 

853 

Whirling of Loaded, 476 

Shape of Stream from Orifice. 534 

of Teeth, 572 

of Worm Thread, 574 

Sharp-edged Notch, 536 

Orifices, 534 

Shear and Twist, 332 

Strain, 882, 341, 350 

Stress, 382 

Stress in Beams, 460 

Shearing Force, 128, 383, 385, 631 

Force in Beams, 414, 427 

Machines, 335 

Ship, 496 

Displacement of a, 55 

Heeling Angle of, 260 

Righting Moment of, 260 

Ships, Friction ot, 56 

Horse-power of, 54 

Models of, 55 

Propulsion of, 52, 525 

Resistance to Motion of, 54 

Speed of, 53 

Shot, Momentum of, 486 
Silicon, 284 

Bronze, 289 

Silver, Alloys of, 289 
Similar Figures, Area of, 8 

Structures, Similarly Loaded, 42-1 

Simple Harmonic Motion, 20, 222. 546 

Harmonic Motion, Amiditude of, 20' 

Harmonic Motion, Frequency of, 20 

Harmonic Motion, Periodic Time of,- 

20 



676 



APPLIED MECHANICS. 



The References are to Pages. 

Simple Harmonic Motions, Combination 

of, 555 
Simpson's Rule, 7 
Sine of an Angle, 23 
Sines, Curve of, 78 
Skeleton Drawings, 221, 570 
Skew Bevel Wheels, 572 
Slider Crank Chain, 577 
Sliding, 106 

■ Contact, 571 

Slip of Belt, 86, 232 
Slope of a Curve, 15 
Smith, Prof. R. H., 45, 582 
Snatch Block, 102 
Soapy "Water, Use of, 237 
Space Average of a Force, 45 

Average, Time Average, 272, 274 

■ Rate, Time Rate, 269 

Specific Gravity, 10 
Speed of Bullet, 14 

of Commercial Shins. 53 

of Train, 12 

Sperm Oil, 66 
Sphere, Surface of, 8 

Volume of, 8 

Spherical Shell, Thick, 325 
Spinning, 106 

Tops, 498 

Spiral Flow of Water, 539, 540 

Spring, Vibration of, 550 

Spring Balance, 40 

of Indiarubber, 41 

of Steel, 41 

Spiral. Vibration of, 550 

Springs, 613 

. Buffer Stop, 613 

C, 643 

Carriage, 618, 646 

■ Clock, 613 

Cylindrical Spiral, 624, 633 

Different Forms of, 618 

Elongation of, 630 

Flat Spiral, 624, 627, 628, 629 

Hardening and Tempering of, 622, 649 

Materials Used in, 620 

Phosphor Bronze, 622 

Resilience of, 620, 641 

^ Spiral, 622 

Tubular Spiral, 642, 643 

Uses of, 620 

Vibration, 615 

which Bend, 462 

Spruce, Norway, 277 
Spur Gearing, 95, 102 

Wheel, 36 

Squared Paper, 1, 13, 15, 21, 44, 60, 89, 90, 

229, 293, 300, 565 
Square-threaded Screw, EfUciency of, 109 
Statical Friction, 75 
Steadiness of Machines, 252 
Steady Motion in Fluids, 532 
Steam Engine, 43 

Turbine, 531 

Steel, 285 
Basic, 286 



The References are to Paats 

Steel, Bessemer, 285 

Cast, 183 

Castings, Annealing of, 2S6 

Crucible, 286 

Hardening of, 290 

Ingots, 285 

Martin, 289 

Rails, 286 

Shear, 285 

Siemens', 286 

Spring, 41 

Strength of, 286 

Tempering of, 28 

Weight of, 9 

Step, 68 

Stepped Cones, 26 
Stiffness of Beams, 432 
Stilling of Vibrations, 565 
Stone, 275 

Artificial, 276 

Stoppage (Sudden) of Water in a Pipe, 

192, 501 
Storage of Energy in Fluids, 321 
Stores of Energy, 42, 509 
Straight Line Law, 62 
Strain, 293, 296. 312 

Energy, 85, 307, 489 

in Castings, 183 

Nature of, 311 

Potential, 362 

Shear, 331, 341 

Strained Spring, Energy of, 41 
Stream Lines, Circular, 541 
Strength of Boiler, 319 

of Chains, 315 

Moilnlns of Sections, 399 

of Pipe, 182, 319 

of Ropes, 315 

Stress, 296 

Principal, 354 

Shear, 331 

Structures, Determination of Forces in, 

156 

Method of Sections, 159 

Struts, 151, 463 

Bending of, 464 

Euler's'Formula, 465 

Rules for, 467 

Shortening of, 298 

with Lateral Loads, 470 

St. Venant, 296, 318, 348, 366, 370, 37Q 
Suddenly Applied Load, 306 
Sulphur, 284 

Sun and Planet Motion, 38 
Surface, Level, 22 

of Fi acture, 306 

Scraped, 73 

Surfaces, Eqni-potcntial, 218 
Suspension, Axis of, 560, 561 

Bifilar, 564 

Bridse, 161 

Switchback Railvray, 41 
Syphon, 79 

Lubricator, 83 

Mr. James Ferry's, 517 



INDEX. 



077 



The References are to Pages. 

T 

Table of Constants, 654 

I. of Sections, 397 

of Strength Moduli, 397 

Tables, Mathematical, 24 

Tangent of an Angle, 23 

Tangential and Normal Components, 106 

Teak, 278 

Teeth of Spur-wheel, 572 

of Wheels, 423 

of Worm-wheel, 572 

Telegraph Wire, Dip of, 168 
Tempering Colours, 291, 292 

of Steel, 290 

Template, 218 
Testing Machine, I 

Machine, Bailey's, 294 

of Water-tube Boiler, 93 

Theorem of Three Moments. 454 
Theory of Bending, 318 

of Fiuid Friction, 80 

Thick Cylinder, 183, 321, 328, 367 

Spherical Shell, 325 

Thin Cylinder, 324 
Thomson's Jet Pump, 516 

Prof. J., Dynamometer, 235 

Prof. J., Overtwisted Shaft, 290 

Triangular Notch, 514, 536 

Turbine, 526 

Whirlpool Chamber, 522 

Thorneycroft's Dynamometer, 234 
Thread, Whitworth, 100 
Threads, Shapes of, 101 
Thurston Oil Tester, 81 
Tide-predicting Machine, Kelvin, 555 
Tide, Kise and Fall of, 554 
Tie Rod, 151 
Tilt Hammer, 500 
Timber, Felling of, 278 

Preserving of, 279 

Seasoning of, 277 

Strength of, 279 

Time Average of a Force, 45 

Time Average, Space Average, 272, 274 

Time, Periodic, 546 

Time Rate, Space Rate, 269 

Tin, 289 

Top, Precession of, 596 

Torque, 77, 95, 127, 260, 351 

Torsional Rigidity, 634 

of a Shaft, 352 

Tortuosity of Path, 606 

Total Energy of 1 lb. of Water, 515, 533 

Toughened Cast Iron, 284 

Glass, 280 

Tower, Mr. Beauchamp, Experiments of, 

73, 81 
Train, Moving, 44 

E pi cyclic, 35 

• Speed of, 12 

Tram car, Pull on, 44 

■ Work Done on, 38 

Trammels, Elliptic, 577 
Translation, Motion of, 121 



The References are to Pagcx 

Transmission Dynamometers, 92, 233 

of Power by Chains, 233 

of Power by Shafts, 224 

Travelling Loads, 436 
Tresca, M., Experiments by, 301 
Triangle of Forces, 56 
Triangular Oriiice, 534 
Trochoidal Curves, 573 
Tube, Built up, 323 

Strengthening of, 324 

Tungsten, 287 
Turbines, 515, 526 

Arranging of, 530 

Axial Mow, 531 

■ Efficiency of, 43 

Guide Blades, 529 

Hydraulic Efficiency of, 629 

Inward Radial Flow, 531 

Outward Radial Flow, 531 

Regulation of, 529 

Steam, 531 

Tweddell's Hydraulic Tools, 199 
Twist, Angle of, 349-352 

and Shear, 332 

Twisting, 349 

and Bending in Shafts, 353 

in Wire, 349 

Moment, 350, 352, 353 

of Prism, 369 



u 

Unclosed Polygon, 124, 129 
Unit of Heat Energy, 142 
Useful Work, 39 
U-Tube, Motion in, 553 



Valve Motions, 584 
Vanes, Radial, 521 

of Centrifugal Pump, 508 

Vector, 29, 123 

Clinure of, 29 

Resolved Part of, 29 

Sense of, 29 

Vectors, Difference of, 30 

Sum of, 30 

Vehicle, Pulling Force on, 108 
Velocity, 12, 14 

Angular, 25 

at Any Instant, 12 

Average, 12 

Critical, 83 

Ratio, 59, 103, 220 

Relative, 32, 34, 35 

of Rubbing, 74 

Venant, St., 296, 318, 348, 366, 370, 379 

Vertical Line, 2* 

Vibrating Adjustable Masses, 501 

Weight, 41 

Vibration, Amplitude of, 565 
Damping of, 566 



67S 



APPLIED MECHANICS 



The R^erejuxs are to Pages. 

Vibration Indicator, 590 

of Hea^y Disc, 56S 

of Spiral Spring, 550 

of Strip of Steel, 552 

Stilling of, 565 

Viscosities, Relative, 569 
Viscosity, 41 

Co^flacient of, 76 

Volts, 92 

Volumes of Solids, SS 

Voussoirs, 163, 165 



W 

Waste of Hydraulic Power, 53 
WaTctanaker's Drill, 291 
Water Flowing Spirally. 539 

Lake of, 512 

Pressure Engine, 1S9. 191 

Pi-essure Machinery, 83 

Pressure of, 167 

Waterfall, Energy of, 59 

'• Waterwitcli " S'teamship, 486 

Watt's Parallel 3Iotion, 69 

Sun and Planet Motion, 36 

Wave Propagation, 367 
Weighbridge, 117 

Graduation of Lever of, 1 18 

Weights of Materials, 9 
Wertheim's Experiments, 302 
Wheels and Axle, 101 

and Axle, Compound, 102 

Friction, 69 

Teeth of. 423 

Teeth, Shapes of. 221 

Value of Train of, 102 

Worm, 102 

Whirling Fluids, 216 

Loaded Shaft, 476 

Whirlpools, 518 

Chamber, Thomson's, 522 

White Fir, 277 

Metal, 75, 289 

Whitworth Thread, 100 



The EeferencfS are to Pages, 

Willans' Engine, 611 
Wind Pressure, 156 
Windmills, 101 
Wire, Area of Section, 10 

Copper, 289 

Drawing Through Die. 299 

Experiment on, 293, 300 

Gun, .^25 

Pianoforte, 2S9 

Ropes, 232 

Telegraph, 168 

Telephone, 289 

Wohler's Experiments, 305. 30s* 
Wooden Pipe. 183 
Work, 38 

by Expanding Fluid. 19 

Cutting of Metals, 46 

Done by a Moment, US 

Done by Turbine, 521 

Law of, 58 

Lost in Friction, 67 

Princiole of, 120 

Useful; 39 

Workshop, 1 

Worm and Worm Wheel, 102, 221 

Thread, 574 

Wrought Iron, 284 

Case-hardening of, 285 

Charcoal, 2S4 

Forging of, 2S4 

— - Lowmoor, 2S4 

Red-short, 2S4 

Staffordshire, 2S4 

Weight of, 9 



Field Point, 300 

Young's Modulus, 299, 305 



Zinc, 287, 283 



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